\(a,\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{2}{y}=2\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{5}{y}=3\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{5}{3}\\\dfrac{2}{x}+\dfrac{9}{5}=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{8}\\y=-\dfrac{5}{3}\end{matrix}\right.\)

\(b,\Leftrightarrow\left\{{}\begin{matrix}\dfrac{60}{x}-\dfrac{28}{y}=36\\\dfrac{60}{x}-\dfrac{135}{y}=525\end{matrix}\right.\left(x,y\ne0\right)\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}+\dfrac{9}{y}=35\\-\dfrac{163}{y}=489\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{4}{x}-27=35\\y=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{31}\\y=-\dfrac{1}{3}\end{matrix}\right.\)

a: Ta có: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{x}-\dfrac{2}{y}=2\\\dfrac{2}{x}-\dfrac{3}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=-3\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-1}{3}\\\dfrac{1}{x}=1+\dfrac{1}{y}=1+\left(-3\right)=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

\(a,ĐK:x\ge0\\ PT\Leftrightarrow4\sqrt{x}-2\sqrt{x}+3\sqrt{x}=12\\ \Leftrightarrow5\sqrt{x}=12\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\\ \Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\\ b,PT\Leftrightarrow\sqrt{\left(2x-3\right)^2}=7\Leftrightarrow\left|2x-3\right|=7\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=7\\3-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

a. 3(x-2)-10=5(2x + 1)

<=> 3x - 6 - 10 = 10x + 5

<=> 3x - 10x = 5 + 6 + 10

<=> -7x = 21

<=> x = -3

b. 3x + 2=8 -2(x-7)

<=> 3x + 2 = 8 - 2x + 14

<=> 3x + 2x = 8 + 14 - 2

<=> 5x = 20

<=> x = 4

c. 2x-(2+5x)= 4(x + 3)

<=> 2x - 2 - 5x = 4x + 12

<=> 2x - 5x - 4x = 12 + 2

<=> -7x = 14

<=> x = -2

d. 5-(x +8)=3x + 3(x-9)

<=> 5 - x - 8 = 3x + 3x - 27

<=> -x - 3x - 3x = -27 + 8 - 5

<=> -7x = -24

<=> x = 24/7

e. 3x - 18 + x= 12-(5x + 3)

<=> 3x - 18 + x = 12 - 5x - 3

<=> 3x + x - 5x = 12 - 3 + 18

<=> -x = 27

<=> x = - 27

Bài 9:

Không, vì $x+2=0$ có nghiệm duy nhất $x=-2$ còn $\frac{x}{x+2}=0$ ngay từ đầu đkxđ đã là $x\neq -2$ (cả 2 pt không có cùng tập nghiệm)

Bài 8:

a. Khi $m=2$ thì pt trở thành:

$(2^2-9)x-3=2$

$\Leftrightarrow -5x-3=2$

$\Leftrightarrow -5x=5$

$\Leftrightarrow x=-1$ 

b.

Khi $m=3$ thì pt trở thành:

$(3^2-9)x-3=3$

$\Leftrightarrow 0x-3=3$

$\Leftrightarrow 0=6$ (vô lý)

c. Khi $m=3$ thì pt trở thành:

$[(-3)^2-9]x-3=-3$

$\Leftrightarrow 0x-3=-3$ (luôn đúng với mọi $x\in\mathbb{R}$)

Vậy pt vô số nghiệm thực.

a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1

=>1,7x=6,7

hay x=67/17

b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)

=>150x+120-45x-75=96x+216-40x+360

=>105x+45=56x+576

=>49x=531

hay x=531/49

a) Ta có: \(\left|x^2-x+2\right|-3x-7=0\)

\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)

\(\Leftrightarrow x^2-x+2=3x+7\)(Vì \(x^2-x+2>0\forall x\))

\(\Leftrightarrow x^2-x+2-3x-7=0\)

\(\Leftrightarrow x^2-4x-5=0\)

\(\Leftrightarrow x^2-5x+x-5=0\)

\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)

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

Vậy: S={5;-1}

a) \(x^2-4x+4=25\\ \Rightarrow\left(x-2\right)^2=25\\ \Rightarrow\left[{}\begin{matrix}x-2=-5\\x-2=5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)

b) \(\left(5-2x\right)^2-16=0\\ \Rightarrow\left(5-2x\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}5-2x=-4\\5-2x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4,5\\0,5\end{matrix}\right.\)

c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\\ \Rightarrow\left(x-3\right)^3-\left(x-3\right)^3+9\left(x+1\right)^2=15\\ \Rightarrow9\left(x+1\right)^2=15\\ \Rightarrow\left(x+1\right)^2=\dfrac{5}{3}\\ \Rightarrow\left[{}\begin{matrix}x+1=-\sqrt{\dfrac{5}{3}}\\x+1=\sqrt{\dfrac{5}{3}}\end{matrix}\right.\)

   \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3+\sqrt{15}}{3}\\x=\dfrac{-3+\sqrt{15}}{3}\end{matrix}\right.\)

a)\(\Leftrightarrow\)\(x^2-4x-21=0\)

\(\Leftrightarrow\)\(x^2-7x+3x-21=0\)

\(\Leftrightarrow\)\(x(x-7)+3(x-7)=0\)

\(\Leftrightarrow\)\((x-7)(x+3)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=7\\ x=-3 \end{array} \right.\)

b)\(\Leftrightarrow\)\((5-2x)^2-4^2=0\)

\(\Leftrightarrow\)\((5-2x-4)(5-2x+4)=0\)

\(\Leftrightarrow\)\((-2x+1)(-2x+9)=0\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=\dfrac{1}{2}\\ x=\dfrac{9}{2} \end{array} \right.\)

c)\((x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2=15\)

\(\Leftrightarrow\)\(x^3-9x^2+27x-27-x^3+27+9x^2+18x+9-15=0\)

\(\Leftrightarrow\)\(45x-6=0\)

\(\Leftrightarrow\)\(x=\dfrac{2}{15}\)

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

a) Áp dụng bđt AM-GM có:

\(\sqrt[3]{\left(9-x\right).8.8}\le\dfrac{9-x+8+8}{3}=\dfrac{25-x}{3}\)\(\Leftrightarrow\sqrt[3]{9-x}\le\dfrac{25-x}{12}\)

\(\sqrt[3]{\left(7+x\right).8.8}\le\dfrac{7+x+8+8}{3}=\dfrac{23+x}{3}\)\(\Leftrightarrow\sqrt[3]{7+x}\le\dfrac{23+x}{12}\)

Cộng vế với vế \(\Rightarrow\sqrt[3]{9-x}+\sqrt[3]{7+x}\le4\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}9-x=8\\7+x=8\end{matrix}\right.\)\(\Rightarrow x=1\)

Vậy...

b)Đk:\(x\ge2\)

Pt \(\Leftrightarrow\left(x-1\right)^2.\left(x^2-4\right)=\left(x-2\right)^2.\left(x^2-1\right)\)

\(\Leftrightarrow\left(x-1\right)^2\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\left(x-1\right)\)

Do \(x\ge2\Rightarrow x-1>0\)

Chia cả hai vế của pt cho x-1 ta được:

\(\left(x-1\right)\left(x-2\right)\left(x+2\right)=\left(x-2\right)^2\left(x+1\right)\)

\(\Leftrightarrow\left(x-2\right)\left[\left(x-1\right)\left(x+2\right)-\left(x-2\right)\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left[x^2+x-2-x^2+3x-2\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)

Vậy S={2}

c)Đk:\(\left\{{}\begin{matrix}9-x^2\ge0\\x^2-1\ge0\\x-3\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}-3\le x\le3\\\left[{}\begin{matrix}x\ge1\\x\le-1\end{matrix}\right.\\x\ge3\end{matrix}\right.\)\(\Rightarrow x=3\)

Thay x=3 vào pt thấy thỏa mãn

Vậy S={3}

a: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=1\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{2}\)

\(a,ĐK:x\ge\dfrac{3}{2}\\ PT\Leftrightarrow\sqrt{2x-3}\left(\sqrt{2x+3}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=0\\\sqrt{2x+3}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{3}{2}\)

\(b,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=1\)