K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
17 tháng 8
Ta có: \(\hat{xAB}=\hat{yBA}\left(=60^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên Ax//By
7 tháng 9
Bài 6: Số học sinh giỏi là \(48\cdot\frac16=8\) (bạn)
Số học sinh trung bình là \(48\cdot25\%=12\) (bạn)
Số học sinh khá là 48-8-12=40-12=28(bạn)
Bài 5:
Thể tích xăng còn lại chiếm:
\(100\%-\frac{3}{10}-40\%=60\%-30\%=30\%\) (tổng số xăng)
Thể tích xăng còn lại là:
\(60\cdot30\%=18\left(lít\right)\)
6 tháng 9
a: \(\left(-\frac54x+3,25\right)\left\lbrack\frac35-\left(-\frac52x\right)\right\rbrack=0\)
=>\(\left(\frac54x-\frac{13}{4}\right)\left(\frac52x+\frac35\right)=0\)
=>\(\left[\begin{array}{l}\frac54x-\frac{13}{4}=0\\ \frac52x+\frac35=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac54x=\frac{13}{4}\\ \frac52x=-\frac35\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:\frac54=\frac{13}{5}\\ x=-\frac35:\frac52=-\frac{6}{25}\end{array}\right.\)
b: \(\left(-\frac72x+1,75\right)\left\lbrack\frac45-\left(-\frac53x\right)\right\rbrack=0\)
=>\(\left[\begin{array}{l}-\frac72x+1,75=0\\ \frac45-\left(-\frac53x\right)=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}-\frac72x=-1,75=-\frac74\\ \frac53x=-\frac45\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{-7}{4}:\frac{-7}{2}=\frac24=\frac12\\ x=-\frac45:\frac53=-\frac45\cdot\frac35=-\frac{12}{25}\end{array}\right.\)
c: \(\left(x^2-4\right)\left(x+\frac27\right)=0\)
=>\(\left[\begin{array}{l}x^2-4=0\\ x+\frac27=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=4\\ x=-\frac27\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-2\\ x=-\frac27\end{array}\right.\)
d: \(\left(25-x^2\right)\left(5x-\frac59\right)=0\)
=>\(\left[\begin{array}{l}25-x^2=0\\ 5x-\frac59=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=25\\ 5x=\frac59\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-5\\ x=\frac19\end{array}\right.\)
13 tháng 8
a: Ta có: \(\hat{AOD}+\hat{BOD}=180^0\) (hai góc kề bù)
=>\(\hat{BOD}=180^0-97^0=83^0\)
Trên cùng một nửa mặt phẳng bờ chứa tia OA, ta có: \(\hat{AOE}<\hat{AOD}\left(56^0<97^0\right)\)
nên tia OE nằm giữa hai tia OA và OD
=>\(\hat{AOE}+\hat{EOD}=\hat{AOD}\)
=>\(\hat{EOD}=97^0-56^0=41^0\)
Ta có: \(\hat{AOE}+\hat{EOC}+\hat{COB}=180^0\)
=>\(\hat{EOC}=180^0-56^0-42^0=82^0\)
b: Trên cùng một nửa mặt phẳng bờ chứa tia OE, ta có; \(\hat{EOD}<\hat{EOC}\left(41^0<82^0\right)\)
nên tia OD nằm giữa hai tia OE và OC
=>\(\hat{EOD}+\hat{DOC}=\hat{EOC}\)
=>\(\hat{DOC}=82^0-41^0=41^0\)
Ta có: tia OD nằm giữa hai tia OE và OC
\(\hat{DOE}=\hat{DOC}\left(=41^0\right)\)
Do đó: OD là phân giác của góc EOC
6 tháng 9
a: Ta có: \(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)
=>\(3x+x-\frac{9}{20}=-\frac{13}{40}\)
=>\(4x=-\frac{13}{40}+\frac{9}{20}=-\frac{13}{40}+\frac{18}{40}=\frac{5}{40}=\frac18\)
=>\(x=\frac18:4=\frac{1}{32}\)
b: \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)
=>\(x+\frac14x-2,5=-\frac{11}{20}\)
=>\(1,25x=-0,55+2,5=1,95\)
=>\(x=\frac{1.95}{1.25}=\frac{195}{125}=\frac{39}{25}\)
c: \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)
=>\(\frac35x+x+0,5=-\frac{13}{15}\)
=>\(\frac85x=-\frac{13}{15}-0,5=-\frac{26}{30}-\frac{15}{30}=-\frac{41}{30}\)
=>\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=\frac{-41}{6\cdot8}=-\frac{41}{48}\)
d: \(-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)
=>\(-\frac23x+4x-\frac67=\frac37\)
=>\(\frac{10}{3}x=\frac37+\frac67=\frac97\)
=>\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)
S
subjects
CTVHS
6 tháng 9
bài 11: câu a:
\(3x+\left(x-\frac{9}{20}\right)=-\frac{13}{40}\)
\(3x+x-\frac{9}{20}=-\frac{13}{40}\)
\(4x=-\frac{13}{40}+\frac{9}{20}\)
\(4x=-\frac{13}{40}+\frac{18}{40}\)
\(4x=\frac{5}{40}\)
\(4x=\frac18\)
\(x=\frac18:4=\frac18\cdot\frac14=\frac{1}{32}\)
b. \(x+\left(\frac14x-2,5\right)=-\frac{11}{20}\)
\(x+\frac14x-2,5=-\frac{11}{20}\)
\(\frac54x-2,5=-\frac{11}{20}\)
\(\frac54x=-\frac{11}{20}+2,5\)
\(\frac54x=\frac{39}{20}\)
\(x=\frac{39}{20}:\frac54=\frac{39}{20}\cdot\frac45=\frac{39}{25}\)
c. \(\frac35x+\left(x+0,5\right)=-\frac{13}{15}\)
\(\frac35x+x+0,5=-\frac{13}{15}\)
\(\frac85x+\frac12=-\frac{13}{15}\)
\(\frac85x=-\frac{13}{15}-\frac12\)
\(\frac85x=-\frac{41}{30}\)
\(x=-\frac{41}{30}:\frac85=-\frac{41}{30}\cdot\frac58=-\frac{41}{48}\)
\(d.-\frac23x+\left(4x-\frac67\right)=\frac{9}{21}\)
\(-\frac23x+4x-\frac67=\frac{9}{21}\)
\(\frac{10}{3}x=\frac97\)
\(x=\frac97:\frac{10}{3}=\frac97\cdot\frac{3}{10}=\frac{27}{70}\)
S
subjects
CTVHS
16 tháng 8
1.1) a) \(\left|2x-5\right|=4\)
\(\Rightarrow\left[\begin{array}{l}2x-5=4\\ 2x-5=-4\end{array}\Rightarrow\left[\begin{array}{l}2x=9\\ 2x=1\end{array}\Rightarrow\left[\begin{array}{l}x=\frac92\\ x=\frac12\end{array}\right.\right.\right.\)
vậy \(x\in\left\lbrace\frac92;\frac12\right\rbrace\)
b)) \(\frac13-\left|\frac54-2x\right|=\frac14\)
\(\left|\frac54-2x\right|=\frac13-\frac14\)
\(\left|\frac54-2x\right|=\frac{1}{12}\)
\(\Rightarrow\left[\begin{array}{l}\frac54-2x=\frac{1}{12}\\ \frac54-2x=-\frac{1}{12}\end{array}\Rightarrow\left[\begin{array}{l}2x=\frac54-\frac{1}{12}\\ 2x=\frac54-\left(-\frac{1}{12}\right)\end{array}\right.\right.\)
\(\Rightarrow\left[\begin{array}{l}2x=\frac76\\ 2x=\frac43\end{array}\Rightarrow\left[\begin{array}{l}x=\frac{7}{12}\\ x=\frac23\end{array}\right.\right.\)
vậy \(x\in\left\lbrace\frac{7}{12};\frac23\right\rbrace\)
\(c.\frac12-\left|x+\frac15\right|=\frac13\)
\(\left|x+\frac15\right|=\frac12-\frac13\)
\(\left|x+\frac15\right|=\frac16\)
\(\Rightarrow\left[\begin{array}{l}x+\frac15=\frac16\\ x+\frac15=-\frac16\end{array}\Rightarrow\left[\begin{array}{l}x=\frac16-\frac15\\ x=-\frac16-\frac15\end{array}\right.\right.\Rightarrow\left[\begin{array}{l}x=-\frac{1}{30}\\ x=-\frac{11}{30}\end{array}\right.\)
vậy \(x\in\left\lbrace-\frac{1}{30};-\frac{11}{30}\right\rbrace\)
\(d.\frac34-\left|2x+1\right|=\frac78\)
\(\left|2x+1\right|=\frac34-\frac78\)
\(\left|2x+1\right|=-\frac18\)
\(\) ⇒ x thuộc rỗng
1.2) a) \(2\left|2x-3\right|=\frac12\)
\(\left|2x-3\right|=\frac12:2=\frac12\cdot\frac12=\frac14\)
\(\left[\begin{array}{l}2x-3=\frac14\\ 2x-3=-\frac14\end{array}\Rightarrow\left[\begin{array}{l}2x=\frac14+3\\ 2x=-\frac14+3\end{array}\right.\right.\)
\(\left[\begin{array}{l}2x=\frac{13}{4}\\ 2x=\frac{11}{4}\end{array}\Rightarrow\left[\begin{array}{l}x=\frac{13}{4}:2=\frac{13}{4}\cdot\frac12=\frac{13}{8}\\ x=\frac{11}{4}:2=\frac{11}{4}\cdot\frac12=\frac{11}{8}\end{array}\right.\right.\)
vậy: \(x\in\left\lbrace\frac{13}{8};\frac{11}{8}\right\rbrace\)
\(\frac{b)1}{3}-\left|\frac54-2x\right|=\frac14\)
\(\left|\frac54-2x\right|=\frac13-\frac14\)
\(\left|\frac54-2x\right|=\frac{1}{12}\)
\(\left[\begin{array}{l}\frac54-2x=\frac{1}{12}\\ \frac54-2x=-\frac{1}{12}\end{array}\Rightarrow\left[\begin{array}{l}2x=\frac54-\frac{1}{12}\\ 2x=\frac54-\left(-\frac{1}{12}\right)\end{array}\right.\right.\)
\(\left[\begin{array}{l}2x=\frac76\\ 2x=\frac43\end{array}\Rightarrow\left[\begin{array}{l}x=\frac76:2=\frac76\cdot\frac12=\frac{7}{12}\\ x=\frac43:2=\frac43\cdot\frac12=\frac23\end{array}\right.\right.\)
vậy \(x\in\left\lbrace\frac{7}{12};\frac23\right\rbrace\)
\(c.\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\left|x+\frac{4}{15}\right|-3,75=-2,15\)
\(\left|x+\frac{4}{15}\right|=3,75-2,15\)
\(\left|x+\frac{4}{15}\right|=1,6\)
\(\left[\begin{array}{l}x+\frac{4}{15}=1,6\\ x+\frac{4}{15}=-1,6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1,6-\frac{4}{15}\\ x=-1,6-\frac{4}{15}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=-\frac{28}{15}\end{array}\right.\)
vậy \(x\in\left\lbrace\frac43;-\frac{28}{15}\right\rbrace\)
16 tháng 8
Bài 1.5:
a: Ta có: \(6,5-\frac94:\left|x+\frac13\right|=2\)
=>\(\frac94:\left|x+\frac13\right|=6,5-2=4,5=\frac92\)
=>\(\left|x+\frac13\right|=\frac94:\frac92=\frac24=\frac12\)
=>\(\left[\begin{array}{l}x+\frac13=\frac12\\ x+\frac13=-\frac12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12-\frac13=\frac16\\ x=-\frac12-\frac13=-\frac56\end{array}\right.\)
b: Ta có: \(\frac{11}{4}+\frac32:\left|4x-\frac15\right|=\frac72\)
=>\(\frac32:\left|4x-\frac15\right|=\frac72-\frac{11}{4}=\frac{14}{4}-\frac{11}{4}=\frac34\)
=>\(\left|4x-\frac15\right|=\frac32:\frac34=\frac42=2\)
=>\(\left[\begin{array}{l}4x-\frac15=2\\ 4x-\frac15=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}4x=2+\frac15=\frac{11}{5}\\ 4x=-2+\frac15=-\frac95\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{20}\\ x=-\frac{9}{20}\end{array}\right.\)
c: Ta có: \(\frac{15}{4}-2,5:\left|\frac34x+\frac12\right|=3\)
=>\(2,5:\left|\frac34x+\frac12\right|=\frac{15}{4}-3=\frac34\)
=>\(\left|\frac34x+\frac12\right|=\frac52:\frac34=\frac52\cdot\frac43=\frac{20}{6}=\frac{10}{3}\)
=>\(\left[\begin{array}{l}\frac34x+\frac12=\frac{10}{3}\\ \frac34x+\frac12=-\frac{10}{3}\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac34x=\frac{10}{3}-\frac12=\frac{20}{6}-\frac36=\frac{17}{6}\\ \frac34x=-\frac{10}{3}-\frac12=-\frac{20}{6}-\frac36=-\frac{23}{6}\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{17}{6}:\frac34=\frac{17}{6}\cdot\frac43=\frac{68}{18}=\frac{34}{9}\\ x=-\frac{23}{6}:\frac34=-\frac{23}{6}\cdot\frac43=\frac{-92}{18}=-\frac{46}{9}\end{array}\right.\)
d: ta có: \(\frac{21}{5}+3:\left|\frac{x}{4}-\frac23\right|=6\)
=>\(3:\left|\frac{x}{4}-\frac23\right|=6-\frac{21}{5}=\frac{30}{5}-\frac{21}{5}=\frac95\)
=>\(\left|\frac{x}{4}-\frac23\right|=3:\frac95=3\cdot\frac59=\frac53\)
=>\(\left[\begin{array}{l}\frac{x}{4}-\frac23=\frac53\\ \frac{x}{4}-\frac23=-\frac53\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{4}=\frac53+\frac23=\frac73\\ \frac{x}{4}=-\frac53+\frac23=-\frac33=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac73\cdot4=\frac{28}{3}\\ x=-1\cdot4=-4\end{array}\right.\)
Bai 1.4:
a: \(\left|x+\frac14\right|-\frac34=5\%\)
=>\(\left|x+\frac14\right|=5\%+\frac34=\frac{1}{20}+\frac{15}{20}=\frac{16}{20}=\frac45\)
=>\(\left[\begin{array}{l}x+\frac14=\frac45\\ x+\frac14=-\frac45\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac45-\frac14=\frac{16}{20}-\frac{5}{20}=\frac{11}{20}\\ x=-\frac45-\frac14=-\frac{16}{20}-\frac{5}{20}=-\frac{21}{20}\end{array}\right.\)
b: \(2-\left|\frac34x-\frac14\right|=\left|-\frac54\right|\)
=>\(2-\left|\frac34x-\frac14\right|=\frac54\)
=>\(\left|\frac34x-\frac14\right|=2-\frac54=\frac34\)
=>\(\left[\begin{array}{l}\frac34x-\frac14=\frac34\\ \frac34x-\frac14=-\frac34\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac34x=\frac34+\frac14=\frac44=1\\ \frac34x=-\frac34+\frac14=-\frac24=-\frac12\end{array}\right.\)
=>\(\left[\begin{array}{l}x=1:\frac34=\frac43\\ x=-\frac12:\frac34=-\frac12\cdot\frac43=-\frac46=-\frac23\end{array}\right.\)
c: \(\frac32+\frac45\left|x-\frac34\right|=\frac74\)
=>\(\frac45\left|x-\frac34\right|=\frac74-\frac32=\frac74-\frac64=\frac14\)
=>\(\left|x-\frac34\right|=\frac14:\frac45=\frac14\cdot\frac54=\frac{5}{16}\)
=>\(\left[\begin{array}{l}x-\frac34=\frac{5}{16}\\ x-\frac34=-\frac{5}{16}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{5}{16}+\frac34=\frac{5}{16}+\frac{12}{16}=\frac{17}{16}\\ x=-\frac{5}{16}+\frac34=-\frac{5}{16}+\frac{12}{16}=\frac{7}{16}\end{array}\right.\)
d: \(4,5-\frac34\left|\frac12x+\frac53\right|=\frac56\)
=>\(\frac34\left|\frac12x+\frac53\right|=4,5-\frac56=\frac92-\frac56=\frac{27}{6}-\frac56=\frac{22}{6}=\frac{11}{3}\)
=>\(\left|\frac12x+\frac53\right|=\frac{11}{3}:\frac34=\frac{11}{3}\cdot\frac43=\frac{44}{9}\)
=>\(\left[\begin{array}{l}\frac12x+\frac53=\frac{44}{9}\\ \frac12x+\frac53=-\frac{44}{9}\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac12x=\frac{44}{9}-\frac53=\frac{44}{9}-\frac{15}{9}=\frac{29}{9}\\ \frac12x=-\frac{44}{9}-\frac53=-\frac{44}{9}-\frac{15}{9}=-\frac{64}{9}\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac{29}{9}:\frac12=\frac{29}{9}\cdot2=\frac{58}{9}\\ x=-\frac{64}{9}:\frac12=-\frac{64}{9}\cdot2=-\frac{128}{9}\end{array}\right.\)
Bài 1.3:
a: \(2\left|3x-1\right|+1=5\)
=>2|3x-1|=4
=>|3x-1|=2
=>\(\left[\begin{array}{l}3x-1=2\\ 3x-1=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=3\\ 3x=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-\frac13\end{array}\right.\)
b: \(\left|\frac{x}{2}-1\right|=3\)
=>\(\left[\begin{array}{l}\frac{x}{2}-1=3\\ \frac{x}{2}-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac{x}{2}=3+1=4\\ \frac{x}{2}=-3+1=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-4\end{array}\right.\)
c: \(\left|-x+\frac25\right|+\frac12=3.5\)
=>\(\left|x-\frac25\right|=3.5-\frac12=\frac72-\frac12=\frac62=3\)
=>\(\left[\begin{array}{l}x-\frac25=3\\ x-\frac25=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3+\frac25=\frac{17}{5}\\ x=-3+\frac25=-\frac{15}{5}+\frac25=-\frac{13}{5}\end{array}\right.\)
d: \(\left|x-\frac13\right|=2\frac15\)
=>\(\left|x-\frac13\right|=\frac{11}{5}\)
=>\(\left[\begin{array}{l}x-\frac13=\frac{11}{5}\\ x-\frac13=-\frac{11}{5}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{11}{5}+\frac13=\frac{33}{15}+\frac{5}{15}=\frac{38}{15}\\ x=-\frac{11}{5}+\frac13=-\frac{33}{15}+\frac{5}{15}=-\frac{28}{15}\end{array}\right.\)
Bài 1.2:
a: \(2\left|2x-3\right|=\frac12\)
=>\(\left|2x-3\right|=\frac14\)
=>\(\left[\begin{array}{l}2x-3=\frac14\\ 2x-3=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+\frac14=\frac{13}{4}\\ 2x=3-\frac14=\frac{11}{4}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{13}{8}\\ x=\frac{11}{8}\end{array}\right.\)
b: \(7,5-3\left|5-2x\right|=-4.5\)
=>3|2x-5|=7,5+4,5=12
=>|2x-5|=4
=>\(\left[\begin{array}{l}2x-5=4\\ 2x-5=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=9\\ 2x=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac92\\ x=\frac12\end{array}\right.\)
c: \(\left|x+\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2.15\right|\)
=>\(\left|x+\frac{4}{15}\right|=-2,15+3,75=1,6=\frac85\)
=>\(\left[\begin{array}{l}x+\frac{4}{15}=\frac85\\ x+\frac{4}{15}=-\frac85\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac85-\frac{4}{15}=\frac{24}{15}-\frac{4}{15}=\frac{20}{15}=\frac43\\ x=-\frac85-\frac{4}{15}=-\frac{24}{15}-\frac{4}{15}=-\frac{28}{15}\end{array}\right.\)
Bài 1.1:
a: |2x-5|=4
=>
S
subjects
CTVHS
28 tháng 8
bài 2: a. ta có góc ADE = góc ABC (= 45 độ)
mà 2 góc này ở vị trí đồng vị
⇒ DE // BC
b. ta có góc FEC = góc ECB
mà 2 góc này ở vị trí so le trong
⇒ EF // BC
c. vì DE // BC và EF // BC nên DE ≡ EF
⇒ 3 điểm D,E,F thẳng hàng
bài 3:
a. ta có góc CHK = góc CAB = 90 độ
mà 2 góc này ở vị trí đồng vị
⇒ KH // AB
b. ta có góc IKB = góc KBA = 60 độ
mà 2 góc này ở vị trí so le trong
⇒ KI // AB
c. vì KH // AB và KI // AB nên KH ≡ KI
⇒ 3 điểm H,K,I thẳng hàng
b: \(\left(x^2-4x+5\right)-\left(x^2-2x+1\right)=3\)
=>\(x^2-4x+5-x^2+2x-1=3\)
=>-2x+4=3
=>-2x=-1
=>\(x=\frac12\)
c: \(\left(5x^4-3x^2+\frac32x-\frac12\right)+\left(3x^2+5x-5x^4+1\right)=3\)
=>\(5x^4-3x^2+\frac32x-\frac12+3x^2+5x-5x^4+1=3\)
=>\(\frac{13}{2}x+\frac12=3\)
=>\(\frac{13}{2}x=3-\frac12=\frac52\)
=>13x=5
=>\(x=\frac{5}{13}\)
b) (x^2-4x+5)-(x^2-2x+1) = 3
x^2-4x+5-x^2+2x-1 = 3
(x^2-x^2) + (-4x+2x) + (5-1) = 3
-2x+4=3
-2x=-1
x=1/2
c) (5x^4 - 3x^2 + 3/2x - 1/2) + (3x^2 + 5x -5x^4 + 1) = 3
5x^4 - 3x^2 + 3/2x - 1/2 + 3x^2 + 5x - 5x^4 + 1 = 3
(5x^4 - 5x^4) + (- 3x^2 + 3x^2) + 3/2x + (- 1/2 + 1) = 3
3/2x + 1/2 = 3
3/2x = 3-1/2
3/2x = 5/2
x = 5/2 : 3/2
x = 5/3