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NL
1
AD
20 tháng 8 2023
Kẻ Az//Bx//Dy
=> BAD = BAz + DAz = (180o - ABx) + (180o - ADy) = 30o + 60o = 90o
QT
1
18 tháng 9 2023
Em thấy bạn Vuông nói đúng
Để chứng minh điều này, ta có thể chỉ ra trường hợp 2 góc bằng nhau nhưng không đối đỉnh.
Ví dụ:
\(\widehat {{O_1}} = \widehat {{O_2}}\) nhưng hai góc này không đối đỉnh
11 tháng 8
Bài 1:
1: xx'⊥AD
yy'⊥AD
Do đó: xx'//yy'
2:
Cách 1:
xx'//yy'
=>\(\hat{C_1}=\hat{x^{\prime}BC}\) (hai góc so le trong)
=>\(\hat{C_1}=70^0\)
Cách 2:
ta có: \(\hat{x^{\prime}BC}+\hat{xBC}=180^0\) (hai góc kề bù)
=>\(\hat{xBC}=180^0-70^0=110^0\)
Ta có: xx'//yy'
=>\(\hat{xBC}+\hat{C_1}=180^0\) (hai góc trong cùng phía)
=>\(\hat{C_1}=180^0-110^0=70^0\)
Bài 2:
a: \(\hat{ABC}=\hat{n^{\prime}CB}\left(=80^0\right)\)
mà hai góc này là hai góc ở vị trí so le trong
nên mm'//nn'
b: Cách 1:
ta có: \(\hat{xAm}+\hat{mAD}=180^0\) (hai góc kề bù)
=>\(\hat{mAD}=180^0-70^0=110^0\)
Ta có: AB//CD
=>\(\hat{mAD}=\hat{D_1}\) (hai góc so le trong)
=>\(\hat{D_1}=110^0\)
Cách 2:
Ta có: \(\hat{xAm}=\hat{BAD}\) (hai góc đối đỉnh)
mà \(\hat{xAm}=70^0\)
nên \(\hat{BAD}=70^0\)
Ta có: AB//CD
=>\(\hat{BAD}+\hat{D_1}=180^0\) (hai góc trong cùng phía)
=>\(\hat{D_1}=180^0-70^0=110^0\)
QT
1
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
Xét 2 tam giác ABC và MNP có:
AB=MN (gt)
\(\widehat {BAC} = \widehat {NMP}\) (gt)
AC=MP (gt)
Vậy \(\Delta ABC = \Delta MNP\)(c.g.c)
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
=>
KV
25 tháng 1 2024
\(5x=3y\Rightarrow x=\dfrac{3y}{5}\)
Thay \(x=\dfrac{3y}{5}\) vào biểu thức \(x^2-y^2=-4\) ta có:
\(\left(\dfrac{3y}{5}\right)^2-y^2=-4\)
\(\dfrac{9y^2}{25}-y^2=-4\)
\(-\dfrac{16}{25}y^2=-4\)
\(y^2=-\dfrac{4}{\dfrac{-16}{25}}\)
\(y^2=\dfrac{25}{4}\)
\(\Rightarrow y=-\dfrac{5}{2};y=\dfrac{5}{2}\)
*) \(y=-\dfrac{5}{2}\Rightarrow x=\dfrac{3.\left(-\dfrac{5}{2}\right)}{5}=-\dfrac{3}{2}\)
*) \(y=\dfrac{5}{2}\Rightarrow x=\dfrac{3.\dfrac{5}{2}}{5}=\dfrac{3}{2}\)
Vậy ta được các cặp giá trị \(\left(x;y\right)\) thỏa mãn:
\(\left(-\dfrac{3}{2};-\dfrac{5}{2}\right);\left(\dfrac{3}{2};\dfrac{5}{2}\right)\)
a) \(\left(x+5\right)^2=36\)
\(\Rightarrow\left(x+5\right)^2=6^2\)
\(\Rightarrow\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6-5\\x=-6-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
b) \(\left(2x-5\right)^3=64\)
\(\Rightarrow\left(2x-5\right)^3=4^3\)
\(\Rightarrow2x-5=4\)
\(\Rightarrow2x=4+5\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
c) \(x^2=2\)
\(\Rightarrow x^2=\left(\sqrt{2}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
d) \(\left(3x-1\right)^2=5\)
\(\Rightarrow\left(3x-1\right)^2=\left(\sqrt{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}3x-1=\sqrt{5}\\3x-1=-\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=1+\sqrt{5}\\3x=1-\sqrt{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{3}\\x=\dfrac{1-\sqrt{5}}{3}\end{matrix}\right.\)
a: =>x+5=6 hoặc x+5=-6
=>x=-11 hoặc x=1
b: (2x-5)^3=64
=>(2x-5)^3=4^3
=>2x-5=4
=>2x=9
=>x=9/2
c: x^2=2
=>\(x^2=\left(\sqrt{2}\right)^2\)
=>\(x=\pm\sqrt{2}\)
d: (3x-1)^2=5
=>\(\left[{}\begin{matrix}3x-1=\sqrt{5}\\3x-1=-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\sqrt{5}+1\\3x=-\sqrt{5}+1\end{matrix}\right.\)
=>\(x=\dfrac{\pm\sqrt{5}+1}{3}\)