Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)\)
<=> \(10x-16-12x+15=12x-16\)
<=> \(10x+15=0\)
<=>x=\(-\frac{15}{10}=-\frac{3}{2}\)
Vậy x=-3/2
a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
\(\text{a) }3x+\dfrac{4}{9}=2x+\dfrac{11}{18}\\ \Leftrightarrow3x-2x=\dfrac{11}{18}-\dfrac{4}{9}\\ \Leftrightarrow x=\dfrac{1}{6}\\ \text{Vậy }x=\dfrac{1}{6}\\ \)
\(\text{b) }\dfrac{7}{12}+\dfrac{2}{3}:x=\dfrac{5}{8}\\ \Leftrightarrow\dfrac{2}{3}:x=\dfrac{1}{24}\\ \Leftrightarrow x=16\\ \text{Vậy }x=16\\ \)
\(\text{c) }\left|2.5-x\right|-\dfrac{1}{5}=1.2\\ \Leftrightarrow\left|2.5-x\right|=1.4\\ \Leftrightarrow\left[{}\begin{matrix}2.5-x=-1.4\\2.5-x=1.4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3.9\\x=1.1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{39}{10}\\x=\dfrac{11}{10}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{39}{10}\text{ hoặc }x=\dfrac{11}{10}\\ \)
\(\text{d) }2^{x+1}+2^{x+2}=192\\ \Leftrightarrow2^x\cdot2+2^x\cdot4=192\\ \Leftrightarrow2^x\left(2+4\right)=192\\ \Leftrightarrow2^x\cdot6=192\\ \Leftrightarrow2^x=32\\ \Leftrightarrow2^x=2^5\\ \Leftrightarrow x=5\\ \text{Vậy }x=5\\ \)
a) x-7 = - 5
<=> x= 2
b) 128 - 3.x.(x+4) = 23
<=> - 3x2 - 12x +105=0
<=> -x2 -4x + 35 =0
\(\Delta'=156\Rightarrow\sqrt{\Delta'}=2\sqrt{39}\)
=> \(\left[{}\begin{matrix}x=-2+\sqrt{39}\\x=-2-\sqrt{39}\end{matrix}\right.\)
vậy S= \(\left\{-2+\sqrt{39};-2-\sqrt{39}\right\}\)
=>3x=3/4+7/5=15/20+28/20=43/20
=>x=43/60