a: =>31-x=60

=>x=-29

b: =>(x-140):35=280-270=10

=>x-140=350

=>x=490

c: =>(1900-2x):35=48

=>1900-2x=1680

=>2x=220

=>x=110

d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)

=>2x-1=11

=>x=6

e: =>(x+2)^5=4^5

=>x+2=4

=>x=2

f: =>3x-4=0 hoặc x-1=0

=>x=4/3 hoặc x=1

g: =>(2x-1)^2=49

=>2x-1=7 hoặc 2x-1=-7

=>x=-3 hoặc x=4

h: =>x(x+1)/2=78

=>x(x+1)=156

=>x=12

a) Ta có: \(3x\left(6x-4\right)-2x\left(9x-1\right)=40\)

\(\Leftrightarrow18x^2-12x-18x^2+2x=40\)

\(\Leftrightarrow-10x=40\)

hay x=-4

Bài 3: 

a: Ta có: 60-3(x-2)=51

\(\Leftrightarrow x-2=3\)

hay x=5

b: Ta có: \(4x-20=25:2^2\)

\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)

hay \(x=\dfrac{105}{16}\)

c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

a.1+3+5+7+...+2x+1=225

[(2x+1-1):2+1]x(2x+1+1):2=225

(x+1)x(2x+2):2=225

(x+1)x(x+1)=225

(x+1)²=225

(x+1)²=15²

x+1=15

x=14

_______________

b. 130-[5.(9-x)+43]=47

    5.(9-x)+43=83

  5.(9-x)=40

9-x=8

x=1

_______________

c.16^x<32⁴

2^4x<2²⁰

=>x∈{0;1;2;3;4}

a: =>(x+1)^2=225

=>x+1=15

=>x=14

b: =>[5*(9-x)+43]=130-47=83

=>5(9-x)=40

=>9-x=8

=>x=1

c: =>2^4x<2^20

=>4x<20

=>0<x<5

a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)

=>\(\left|2x+1\right|=\dfrac{1}{2}\)

=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

c: (2x-3)²=36

=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

d: \(7^{x+2}+2\cdot7^x=357\)

=>\(7^x\cdot49+7^x\cdot2=357\)

=>\(7^x=7\)

=>x=1

a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)

\(---\)

b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)

\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)

\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)

\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)

\(---\)

c) \(\left(2x-3\right)^2=36\)

\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(---\)

d) \(7^{x+2}+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)

\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)

\(\Rightarrow7^x\cdot\left(49+2\right)=357\)

\(\Rightarrow7^x\cdot51=357\)

\(\Rightarrow7^x=357:51\)

\(\Rightarrow7^x=7\)

\(\Rightarrow x=1\)

17x + 3. ( -16x – 37) = 2x + 43 - 4x

<=>17x-48x-111=-2x+43

<=>-29x=154

<=> \(x=-\frac{154}{29}\)

-3. (2x + 5) -16 < -4. (3 – 2x)

\(\Leftrightarrow-6x-31< -12+8x.\)

\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)

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