1, bạn xem lại đề 

2, 15(x-3) + 8x-21 = 12(x+1) +120 

<=> 23x - 66 = 12x + 132 

<=> 11x = 198 <=> x = 198/11 

3, 10(3x+1) + 5 - 100 = 8(3x-1) - 6x - 4 

<=> 30x + 10 - 95 = 18x -12

<=> 12x = 73 <=> x = 73/12 

\(4.3^x+3^{x+1}=63\)

\(\Rightarrow4.3^x+3.3^x=63\)

\(\Rightarrow7.3^x=63\Rightarrow3^x=9=3^2\Rightarrow x=2\)

\(9.\left(\dfrac{2}{3}\right)^{x+2}-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow9.\left(\dfrac{2}{3}\right)^2\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow9.\dfrac{4}{9}^{ }.\left(\dfrac{2}{3}\right)^x-\left(\dfrac{2}{3}\right)^x=\dfrac{4}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\left(4-1\right)=\dfrac{4}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x.\dfrac{1}{3}=\dfrac{4}{3}\Rightarrow\left(\dfrac{2}{3}\right)^x=4\)

mà \(0< \left(\dfrac{2}{3}\right)^x< 1;4>0;x>0\)

\(\Rightarrow x\in\varnothing\)

Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)

$\Rightarrow -11x\geq 0$

$\Rightarrow x\leq 0$

Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$

PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

Lời giải:
Vế trái luôn không âm (tính chất trị tuyệt đối)

$\Rightarrow -11x\geq 0$

$\Rightarrow x\leq 0$

Do đó: $x-\frac{1}{3}, x-\frac{1}{15},..., x-\frac{1}{399}<0$

PT trở thành:
$\frac{1}{3}-x+\frac{1}{15}-x+...+\frac{1}{399}-x=-11x$

$(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{399})-10x=-11x$

$\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{19.21}=-x$

$\frac{1}{2}(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{19}-\frac{1}{21})=-x$

$\frac{1}{2}(1-\frac{1}{21})=-x$

$\frac{10}{21}=-x$

$\Rightarrow x=\frac{-10}{21}$

a) (-5/9)^10 : x = (-5/9)^8

=> x = (-5/9)^10 : (-5/9)^8

=> x = (-5/9)^10-8 = (-5/9)^2

=> x = 25/81

b ) x : (-5/9)^8 = (-9/5)^8

=> x = (-9/5)^8 . (-5/9)^8

=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )

=> x = 1

C) x^3 = -8 =(-2)^3

=> x = -2

a) (-5/9)¹⁰ : x = (-5/9)⁸

x = (-5/9)¹⁰ : (-5/9)⁸

x = (-5/9)²

x = 25/81

b) x : (-5/9)⁸ = (-9/5)⁸

x = (-9/5)⁸ . (-5/9)⁸

x = [-9/5 . (-5/9)]⁸

x = 1⁸

x = 1

c) x³ = -8

x³ = (-2)³

x = -2

\(a,\left|x\right|< 2\dfrac{3}{4}\)

\(\Rightarrow-2\dfrac{3}{4}< x< 2\dfrac{3}{4}\)

mà \(x\in Z\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

\(b,\left|x\right|>1\dfrac{3}{8}\)

\(\Rightarrow-1\dfrac{3}{8}< x< 1\dfrac{3}{8}\)

mà \(x\in Z\)

\(\Rightarrow x\in\left\{;-1;0;1;\right\}\)

\(\)

c: =>7/6<|x-2/3|<26/9

=>7/6<x-2/3<26/9 hoặc -7/6>x-2/3>-26/9

=>11/6<x<32/9 hoặc -1/2>x>-20/9

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

a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)

\(=-\sqrt{7}\)

\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

b) \(A>B\) khi 

\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)

\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)

\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)

bài 2:tính hợp lý

1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)

\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)

\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)

\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)

Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)

Bài 2: 

a: \(=\dfrac{-6\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}{9\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{29}\right)}=\dfrac{-6}{9}=\dfrac{-2}{3}\)

b: \(=\dfrac{\dfrac{2}{15}-\dfrac{2}{21}+\dfrac{2}{39}}{\dfrac{10}{40}-\dfrac{10}{56}+\dfrac{10}{104}}\)

\(=\dfrac{\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}{\dfrac{10}{8}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{2}{3}:\dfrac{5}{4}=\dfrac{2}{3}\cdot\dfrac{4}{5}=\dfrac{8}{15}\)

c: \(=\dfrac{2\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}{4\left(25-\dfrac{2}{13}+\dfrac{1}{15}-\dfrac{1}{17}\right)}:\dfrac{1+\dfrac{2}{21}-\dfrac{5}{121}}{13\left(\dfrac{5}{121}-\dfrac{2}{21}-1\right)}\)

=2/4:(-1)/13=2/4x(-13)=-13/2

câu c) mang tính mua vui hay gì hả bn

mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)

Do \(\left|x-\dfrac{2}{3}\right|\ge0;\forall x\)

Mà \(-\dfrac{26}{\sqrt{81}}< 0\)

\(\Rightarrow\) Không tồn tại x để \(\left|x-\dfrac{2}{3}\right|< -\dfrac{26}{\sqrt{81}}\)

Hay ko tồn tại số nguyên x thỏa mãn đề bài