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a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
Chúc bạn học tốt!
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
b) \(\left(5x-1\right)\left(2x-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\2x-\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=1\\2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{5}\\x=\frac{1}{6}\end{matrix}\right.\)
e, \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}-\left(-1\right)\)
\(\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{15}\\x=1,05\end{matrix}\right.\)
Vậy ....
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
a) \(\left(\frac{5}{7}x-\frac{1}{4}\right)\left(\frac{-3}{4}x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{7}x-\frac{1}{4}=0\\\frac{-3}{4}x+\frac{1}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{7}x=\frac{1}{4}\\\frac{-3}{4}x=\frac{-1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{7}{20}\\x=\frac{2}{3}\end{cases}}\)
Vậy \(x=\frac{7}{20}\) hoặc x=\(\frac{2}{3}\)
b) \(\left(\frac{4}{5}+x\right)\left(x-\frac{8}{13}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{4}{5}+x=0\\x-\frac{8}{13}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-4}{5}\\x=\frac{8}{13}\end{cases}}\)
Vậy x=-4/5 hoặc x=8/13
c) \(\left(2x-\frac{1}{2}\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-\frac{1}{2}=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=3\end{cases}}\)
Vậy x=1/4 hoặc x=3
\(x+\frac{7}{2}x+x=\frac{1}{2}\)
\(2x+\frac{7}{2}x=\frac{1}{2}\)
\(\left(2+\frac{7}{2}\right)x=\frac{1}{2}\)
\(\frac{11}{2}x=\frac{1}{2}\)
\(x=\frac{1}{2}:\frac{11}{2}\)
\(x=\frac{1}{11}\)
a) \(\left|0,5x-2\right|-\left|x+\frac{1}{3}\right|=0\)
=> \(\left|0,5x-2\right|=\left|x+\frac{1}{3}\right|\)
=> \(\orbr{\begin{cases}0,5x-2=x+\frac{1}{3}\\0,5x-2=-x-\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}-0,5x=\frac{7}{3}\\1,5x=\frac{5}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{14}{3}\\x=\frac{10}{9}\end{cases}}\)
b) \(2x-\left|x+1\right|=\frac{1}{2}\)
=> \(\left|x+1\right|=2x-\frac{1}{2}\) (Đk: \(2x-\frac{1}{2}\ge0\) <=> \(x\ge\frac{1}{4}\))
=> \(\orbr{\begin{cases}x+1=2x-\frac{1}{2}\\x+1=\frac{1}{2}-2x\end{cases}}\)
=> \(\orbr{\begin{cases}-x=-\frac{3}{2}\\3x=-\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{6}\end{cases}}\)