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2 tháng 4 2020
1) 
\(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)
\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)
\(x+\frac{4}{5}=\pm\frac{4}{5}\)
\(TH1:x+\frac{4}{5}=\frac{4}{5}\)
\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)
\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)
Vậy x ∈ {0; \(\frac{-8}{5}\)}
Hai câu cuối khó nhìn nên ko giải
TN
0
HN
28 tháng 8 2021
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
HN
28 tháng 8 2021
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
2 tháng 4 2020
1: Ta có: \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x-3x^2+x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow-2x\left(x+1\right)=0\)
Vì -2≠0
nên \(\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: x∈{0;-1}
2) Ta có: \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x-12+x=0\)
\(\Leftrightarrow11x-2=0\)
\(\Leftrightarrow11x=2\)
hay \(x=\frac{2}{11}\)
Vậy: \(x=\frac{2}{11}\)
3) Ta có: \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}-5=0\)
\(\Leftrightarrow\frac{-13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow\frac{4}{3}x=\frac{-13}{3}\)
hay \(x=\frac{-13}{3}:\frac{4}{3}=\frac{-13}{4}\)
Vậy: \(x=\frac{-13}{4}\)
4) Ta có: \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{5}\\x-\frac{4}{5}=\frac{-3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{5}\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{5};\frac{7}{5}\right\}\)
BL
2 tháng 4 2020
1. \(2x+x\left(x-5\right)=3x^2-x\)
\(\Leftrightarrow2x+x^2-5x=3x^2-x\)
\(\Leftrightarrow\left(2x-5x+x\right)+\left(x^2-3x^2\right)=0\)
\(\Leftrightarrow-2x-2x^2=0\)
\(\Leftrightarrow-2x\left(1+x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\1+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
2. \(15-5\left(1-2x\right)=12-x\)
\(\Leftrightarrow15-5+10x=12-x\)
\(\Leftrightarrow\left(15-5-12\right)+\left(10x+x\right)=0\)
\(\Leftrightarrow-2+11x=0\)
\(\Leftrightarrow11x=2\Leftrightarrow x=\frac{2}{11}\)
3. \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
\(\Leftrightarrow\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}-5\right)-\left(\frac{1}{3}x+x\right)=0\)
\(\Leftrightarrow-\frac{13}{3}-\frac{4}{3}x=0\)
\(\Leftrightarrow-\frac{4}{3}x=\frac{13}{3}\Leftrightarrow x=-\frac{13}{4}\)
4. \(\left|x-\frac{4}{5}\right|=\frac{3}{5}\)
\(\Rightarrow x-\frac{4}{5}=-\frac{3}{5}\) hoặc \(x-\frac{4}{5}=\frac{3}{5}\)
\(TH1:x-\frac{4}{5}=-\frac{3}{5}\Rightarrow x=\frac{1}{5}\)
\(TH2:x-\frac{4}{5}=\frac{3}{5}\Rightarrow x=\frac{7}{5}\)
22 tháng 9 2017
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
25 tháng 6 2019
a, (ko vt lại đề)
=> -5x- 1-1/2x -1/3=3/2x -5/6
=> -5x - 1/2x +3/2x = 1+1/3 - 5/6
=>( -5 -1/2 + 3/2 )x =1/2
=> -4x = 1/2
=> x = -1/8
17 tháng 10 2020
a) \(\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
=> \(\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x+\frac{1}{3}=0\)
=> \(\left(\frac{2}{3}x-\frac{1}{2}x\right)+\left(-\frac{2}{5}+\frac{1}{3}\right)=0\)
=> \(\frac{1}{6}x-\frac{1}{15}=0\Rightarrow\frac{1}{6}x=\frac{1}{15}\Rightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{2}{5}\)
Vậy x = 2/5
b) \(\frac{1}{3}x+\frac{2}{5}\left(x+1\right)=0\)
=> \(\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
=> \(\frac{11}{15}x+\frac{2}{5}=0\Rightarrow\frac{11}{15}x=-\frac{2}{5}\)
=> \(x=\left(-\frac{2}{5}\right):\frac{11}{15}=\left(-\frac{2}{5}\right)\cdot\frac{15}{11}=-\frac{6}{11}\)
Vậy x = -6/11
c) \(\frac{2}{3}-\frac{1}{3}\left(x-\frac{3}{2}\right)-\frac{1}{2}\left(2x+1\right)=5\)
=> \(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
=> \(\left(\frac{2}{3}+\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{1}{3}x-x\right)=5\)
=> \(\frac{2}{3}-\frac{4}{3}x=5\)
=> \(\frac{4}{3}x=-\frac{13}{3}\Rightarrow x=\left(-\frac{13}{3}\right):\frac{4}{3}=\left(-\frac{13}{3}\right)\cdot\frac{3}{4}=-\frac{13}{4}\)
Vậy x = -13/4
d) \(\frac{11}{5}-\left(\frac{7}{9}-x\right)\cdot\frac{3}{8}=\frac{61}{90}+\frac{x}{3}\)
=> \(\frac{11}{5}-\frac{3}{8}\left(\frac{7}{9}-x\right)=\frac{61}{90}+\frac{30x}{90}\)
=> \(\frac{11}{5}-\frac{7}{24}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3}{8}x=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{3x}{8}=\frac{61+30x}{90}\)
=> \(\frac{229}{120}+\frac{45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{229+45x}{120}=\frac{61+30x}{90}\)
=> \(\frac{3\left(229+45x\right)}{360}=\frac{4\left(61+30x\right)}{360}\)
=> \(3\left(229+45x\right)=4\left(61+30x\right)\)
=> \(687+135x=244+120x\)
=> \(687+135x-244-120x=0\)
=> \(\left(687-244\right)+\left(135x-120x\right)=0\)
=> \(443+15x=0\)
=> \(15x=-443\Rightarrow x=-\frac{443}{15}\)
Vậy x = -443/15
\(\dfrac{2}{3}.x-1\dfrac{2}{5}.x=\dfrac{3}{5}\)
\(\dfrac{2}{3}.x-\dfrac{7}{5}.x=\dfrac{3}{5}\)
\(x.\left(\dfrac{2}{3}-\dfrac{7}{5}\right)=\dfrac{3}{5}\)
\(x.\left(\dfrac{10}{15}-\dfrac{21}{15}\right)=\dfrac{3}{5}\)
\(x.\dfrac{-11}{15}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}:\dfrac{-11}{15}=\dfrac{3}{5}.\dfrac{-15}{13}\)
\(x=\dfrac{-9}{11}\)