\(x-1\in\left\{1;6;2;3;-1;-6;-2;-3\right\}\)

\(\Leftrightarrow x\in\left\{2;7;3;4;0;-5;-1;-2\right\}\)

\(10⋮2x+1\)

\(\Rightarrow2x+1\in\left\{1;2;5;10;-1;-2;-5;-10\right\}\)

\(\Rightarrow2x\in\left\{0;1;4;9;-2;-6;-11\right\}\)

\(\Leftrightarrow x\in\left\{0;\frac{1}{2};2;\frac{9}{2};-1;-3;-\frac{11}{2}\right\}\)

\(x+13⋮x+1\)

\(\Leftrightarrow\left(x+1\right)+12⋮x+1\)

Do \(x+1⋮x+1\) nên \(12⋮x+1\)

\(\Rightarrow x+1\in\left\{1;12;6;2;4;3;-1;-12;-6;-2;-4;-3\right\}\)

\(\Rightarrow x\in\left\{0;11;5;1;3;2;-2;-13;-7;-3;-5;-4\right\}\)

\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)

Câu a) xem lại đề giùm nhé em

b) \(\left(x-1\right)^3=9^3\)

\(x-1=9\)

\(x=10\)

Vậy \(x=10\)

c) \(\left(x-1\right)^2=25\)

\(x-1=5\) hoặc \(x-1=-5\)

* \(x-1=5\)

\(x=6\)

* \(x-1=-5\)

\(x=-4\)

Vậy \(x=-4\); \(x=6\)

d) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

Vậy \(x=2\)

e) Sửa đề: \(\left(2x+4\right)^3=64\)

\(\left(2x+4\right)^3=4^3\)

\(2x+4=4\)

\(2x=0\)

\(x=0\)

Vậy \(x=0\)

b: =>2/5*x=2/3+4/5=22/15

=>x=11/3

c: =>2,5-0,25(2-1/2x)=0,25

=>0,25(2-0,5x)=2,25

=>2-0,5x=9

=>-0,5x=-7

=>x=14

d: =>(x-3)^2=36

=>x=9 hoặc x=-3

e: =>1/2x-3/4=0 và x+y=25

=>x=15 và y=10

1.A.0.96

Câu a tự làm nhé

b, \(\frac{2x+3}{24}=\frac{3x-1}{32}\)

\(\Leftrightarrow32(2x+3)=24(3x-1)\)

\(\Leftrightarrow64x+96=72x-24\)

\(\Leftrightarrow64x+96-72x=-24\)

\(\Leftrightarrow96-8x=-24\Leftrightarrow x=15\)

2. a. \(\frac{x}{3}=\frac{y}{4}=\frac{z}{6}\) và x + y + z = 52

Ta có : \(\frac{x}{3}=\frac{y}{4}=\frac{z}{6}=\frac{x+y+z}{3+4+6}=\frac{52}{13}=4\)\((\)áp dụng tính chất dãy tỉ số bằng nhau \()\)

Vậy : \(\hept{\begin{cases}\frac{x}{3}=4\\\frac{y}{4}=4\\\frac{z}{6}=4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=12\\y=16\\z=24\end{cases}}\)

Câu 1:

( 2 + 4 + 6 + ... + 40 ) - x = 280

SSH là : ( 40 - 2 ) : 2 + 1 = 20 ( số )

Tổng là : ( 40 + 2 ) . 20 : 2 = 420

=> 420 - x = 280

=> x = 140

P/s : mk làm từng phần một nha

( x + 3 ) + ( x + 5 ) + ( x + 7 ) + ( x + 9 ) = 72

4x + 24 = 72

4x = 48

x = 12

( 3 + 6 + ... + 45 ) - 2x = 40

SSH là : ( 45 - 3 ) : 3 + 1 = 15 ( số )

Tổng là : ( 45 + 3 ) . 15 ; 2 = 360

=> 360 - 2x = 40

=> 2x = 320

=> x = 160

a)\(2x+3\left(x+4\right)-14=8\)

\(2x+3x+12-14=8\)

\(2x+3x-2=8\)

\(5x-2=8\)

\(5x=8+2\)

\(5x=10\)

\(x=10:5\)

\(x=2\)

b)\(x+2x+3x=4x+16\)

\(x\left(1+2+3\right)=4x+16\)

\(6x-4x=16\)

\(2x=16\)

\(x=16:2\)

\(x=8\)

c)\(2\left(x+1\right)+3\left(x+5\right)=0\)

\(2x+2+3x+15=0\)

\(5x+17=0\)

\(5x=0-17\)

\(5x=-17\)

\(x=-17:5\)

\(x=-\frac{17}{5}\)

Câu 2 sai đề nhé 

Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)