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câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a.
| x | = 5,6
=>\(\left[{}\begin{matrix}x=5,6\\x=-5,6\end{matrix}\right.\)
Vậy \(x\in\left\{-5,6;5,6\right\}\)
b, \(\left|x-3,5\right|=5\)
=>\(\left[{}\begin{matrix}x-3,5=5\\x-3,5=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8,5\\x=-1,5\end{matrix}\right.\)
Vậy \(x\in\left\{-1,5;8,5\right\}\)
c,\(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=> \(\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{4};\dfrac{5}{4}\right\}\)
d,\(\left|4x\right|-\left(\left|-13,5\right|\right)=\left|\dfrac{1}{4}\right|\)
=> \(\left|4x\right|-13,5=\dfrac{1}{4}\)
=> \(\left|4x\right|=13,75\)
=>\(\left[{}\begin{matrix}4x=13,75\\4x=-13,75\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3,4375\\x=-3,4375\end{matrix}\right.\)
Vậy \(x\in\left\{-3,4375;3,4375\right\}\)
e, ( x - 1 ) 3 = 27
=> x - 1 = 3
=> x = 4
Vậy x = 4
f, ( 2x - 3)2 = 36
=> \(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=4,5\\x=-1,5\end{matrix}\right.\)
Vậy x\(\in\left\{-1,5;4,5\right\}\)
g, \(5^{x+2}=625\)
=> \(5^{x+2}=5^4\)
=> x + 2 = 4
=> x = 2
Vậy x = 2
h, ( 2x - 1)3 = -8
=> 2x - 1 = -2
=> x = \(\dfrac{-1}{2}\)
Vậy x = \(\dfrac{-1}{2}\)
i, \(\dfrac{1}{4}.\dfrac{2}{6}.\dfrac{3}{8}.\dfrac{4}{10}.\dfrac{5}{12}...\dfrac{30}{62}.\dfrac{31}{64}=2^x\)
=> \(\dfrac{1.2.3.4.5...30.31}{4.6.8.10.12...62.64}=2^x\)
=>\(\dfrac{1.2.3.4.5...30.31}{\left(2.3.4.5...30.31.32\right)\left(2.2.2.2...2.2_{ }\right)}=2^x\)(có 31 số 2)
=> \(\dfrac{1}{32.2^{31}}=2^x\)
=> \(\dfrac{1}{2^{36}}=2^x\)
=> x = -36
Vậy x = -36
a, \(\dfrac{3}{4}+x=\dfrac{8}{13}\)
\(x=\dfrac{8}{13}-\dfrac{3}{4}\)
\(x=-\dfrac{7}{52}\)
b,\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
c, \(2x\left(x-\dfrac{1}{7}\right)=0\)
\(2x-\dfrac{1}{7}=0\)
\(x-\dfrac{1}{7}=0:2\)
\(x-\dfrac{1}{7}=0\)
\(x=0-\dfrac{1}{7}\)
\(x=\dfrac{1}{7}\)
d, \(\dfrac{3}{4}+\dfrac{1}{4}\div x=\dfrac{2}{5}\)
\(\left(\dfrac{3}{4}+\dfrac{1}{4}\right):x=\dfrac{2}{5}\)
\(1:x=\dfrac{2}{5}\)
\(x=1:\dfrac{2}{5}\)
\(x=\dfrac{5}{2}\)
a) \(\dfrac{3}{4}+x=\dfrac{8}{13}\)\(\Leftrightarrow\) \(x=\dfrac{8}{13}-\dfrac{3}{4}=\dfrac{-7}{52}\) vậy \(x=\dfrac{-7}{52}\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\) \(\Leftrightarrow\) \(\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\) \(\Leftrightarrow\) \(x=\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=\dfrac{-3}{20}\) vậy \(x=\dfrac{-3}{20}\)
c) \(2x\left(x-\dfrac{1}{7}\right)=0\) \(\Leftrightarrow\) \(2x^2-\dfrac{2}{7}x=0\)
\(\Delta\) = \(\left(\dfrac{-2}{7}\right)^2-4.2.0=\dfrac{4}{49}>0\)
\(\Rightarrow\) phương trình có 2 nghiệm phân biệt
\(x_1=\dfrac{\dfrac{2}{7}+\sqrt{\dfrac{4}{49}}}{4}=\dfrac{1}{7}\)
\(x_2=\dfrac{\dfrac{2}{7}-\sqrt{\dfrac{4}{49}}}{4}=0\)
vậy \(x=0;x=\dfrac{1}{7}\)
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
a) \(\dfrac{5}{6}:x=30:3\)
\(\Leftrightarrow\dfrac{5}{6}:x=10\)
\(\Leftrightarrow x=\dfrac{5}{6}:10\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy .......
b) \(x:2,5=0,003:0,75\)
\(\Leftrightarrow x:2,5=0,004\)
\(\Leftrightarrow x=0,004.2,5\)
\(\Leftrightarrow x=0,01\)
Vậy .......
c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)
\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)
\(\Leftrightarrow2x=\dfrac{698}{25}\)
\(\Leftrightarrow x=\dfrac{304}{15}\)
Vậy ...
d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)
\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{8}{15}\)
Vậy ....
e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)
\(\Leftrightarrow0,25x=\dfrac{57}{608}\)
\(\Leftrightarrow x=\dfrac{228}{608}\)
Vậy ...
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)
\(\Leftrightarrow x^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy ...
1)
a) \(\frac{x}{6}\)= \(\frac{7}{3}\)
\(\Rightarrow\)x.3=6.7
\(\Rightarrow\)x.3=42
\(\Rightarrow\)x =42:3
\(\Rightarrow\)x =14
b) làm tương tự như câu a
c) làm tương tự như câu
d) làm tương tư như câu a nhưng hơi phúc tạp một chút là bn phải đổi ra từ hỗn số ra phân số hoặc số nguyên
e) tương tự câu d
f) làm tương tự như câu d
2)
a) 3x:\(\frac{27}{10}\)=\(\frac{1}{3}\): \(2\frac{1}{4}\)
3x: \(\frac{27}{10}\) = \(\frac{1}{3}\): \(\frac{9}{4}\)
3x: \(\frac{27}{10}\) = \(\frac{4}{27}\)
3x = \(\frac{4}{27}\). \(\frac{27}{10}\)
3x = \(\frac{2}{5}\)
x = \(\frac{2}{5}\): 3
x = \(\frac{2}{15}\)
Các câu còn lại bn làm tương tự như câu a nha
3)
Làm tương tự như bài 2 nha
mik khuyên bn nếu bn giải bài thì bn nên đổi ra cùng một kiểu số thì tốt hơn như số số thập phân thì thập phân hết ấy
Cuối cùng chúc bn học giỏi
a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)
\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)
=>x=13/12
b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)
\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)
\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)
d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)
\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)
e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)
=>x+2020=0
hay x=-2020
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a,\(x+\frac{1}{4}=\frac{3}{4} \)
<=>x=\(\frac{3}{4}-\frac{1}{4} \)
<=>\(x=\frac{1}{2} \)
b,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60} \)
<=>\(\frac{2}{5}x =\frac{3}{4}-\frac{29}{60} \)
<=>\(\frac{2}{5}x=\frac{4}{15} \)
<=x=\(\frac{2}{3} \)
c,\(2x-\frac{1}{3}=\frac{-5}{6} \)
2x=\(\frac{-5}{6}+\frac{1}{3} \)
2x=\(\frac{-1}{2} \)
x=\(\frac{-1}{4} \)
d,2-\(\frac{3}{4x}=\frac{1}{2} \)
<=>\(\frac{3}{4x}=2-\frac{1}{2} \)
<=>\(\frac{3}{4}x=\frac{3}{2} \)
<=>x=2
e,\(\frac{11}{12}- \frac{2}{3}|x| =\frac{3}{8} \)
<=>\(\frac{2}{3}|x|=\frac{13}{24} \)
<=>\(|x|=\frac{13}{16} \)
<=>x=\(\pm\frac{13}{16} \)
f,\(|2x-1|=5\)
<=>2x-1=5 hoặc 2x-1=-5
<=> 2x=6 2x=-4
<=> x=3 x=-2
Giải
a) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
=> \(x=\dfrac{3}{4}-\dfrac{1}{4}\)=>\(x=\dfrac{1}{2}\)
b)\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}x=\dfrac{3}{4}-\dfrac{29}{60}\)=>\(\dfrac{2}{5}x=\dfrac{4}{15}\)
=>\(x=\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{2}{3}\)
c)\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)=>\(2x=\dfrac{-5}{6}+\dfrac{1}{3}\)=\(\dfrac{-1}{2}\)
=>\(x=\dfrac{-1}{2}:2=\dfrac{-1}{4}\)
d)\(2-x:\dfrac{3}{4}=\dfrac{1}{2}\)=>\(x:\dfrac{3}{4}=2-\dfrac{1}{2}\)=\(\dfrac{3}{2}\)
=>\(x=\dfrac{3}{2}.\dfrac{3}{4}=\dfrac{9}{8}\)
e)\(\dfrac{11}{12}-\dfrac{2}{3}.\left|x\right|=\dfrac{3}{8}\)=>\(\dfrac{2}{3}.\left|x\right|=\dfrac{11}{12}-\dfrac{3}{8}\)=\(\dfrac{13}{24}\)
=>\(\left|x\right|=\dfrac{13}{24}:\dfrac{2}{3}=\dfrac{13}{16}\)
Vậy: \(x=\dfrac{13}{16}\)hoặc\(x=\dfrac{-13}{16}\)
f)\(\left|2x-1\right|=5\)
*2x-1=5 =>2x=5+1=6 =>x=6:2=3
*2x-1=-5 =>2x=(-5)+1=-4 =>x=-4:2=-2
Vậy: x=3 hoặc x=-2
Tick cho Phong nhé:>
Yêu nhiều>3
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