Lời giải:
a.

\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)

$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix} x^2-1=(x-1)(x+1)\neq 0\\ 7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq \pm 1\\ x\leq \frac{7}{2}\end{matrix}\right.\)

c.

\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)

d.

\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)

a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)

Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)

\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)

\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)

h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)

Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)

\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)

Thay a,b:

\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x

=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x

=> 2x+3y-1 / 12 = 2x+3y-1 / 6x

=> 12 = 6x => x =2

a ) \(\left(2x-1\right)^2+\left(x+4\right)^2+5=2x\left(x+1\right)+\left(x+2\right)^2+2x^2-2x+18\)

\(\Leftrightarrow4x^2-4x+1+x^2+8x+16+5=2x^2+2x+x^2+4x+4+2x^2-2x+18\)

\(\Leftrightarrow5x^2+4x+22=5x^2+4x+22\)

=> PT có vô số nghiệm

b ) \(\dfrac{5x-7}{4}-\dfrac{9x-4}{5}=-x-\dfrac{19-9x}{20}\)

\(\Leftrightarrow\dfrac{25x-35-36x+16}{20}=\dfrac{-20x-19+9x}{20}\)

\(\Leftrightarrow\dfrac{-11x-19}{20}=\dfrac{-11x-19}{20}\)

=> PT có vô số nghiệm

c ) \(\left|y-3\right|=y-3\)

TH 1 : \(y\ge3\)

\(\Rightarrow y-3\ge0\Rightarrow\left|y-3\right|=y-3\)

Do \(\left|y-3\right|=y-3\)

\(\Rightarrow y-3=y-3\)

Nên : \(y\ge3\) , PT vô số nghiệm

TH 2 : \(y< 3\Rightarrow y-3< 0\Rightarrow\left|y-3\right|=3-y\)

Do \(\left|y-3\right|=y-3\)

\(\Rightarrow3-y=y-3\)

\(\Rightarrow3-y-y+3=0\)

\(\Rightarrow6-2y=0\)

\(\Rightarrow y=3\) ( L ; do y < 3 )

Vậy \(y\ge3\) thì PT vô số nghiệm

a: \(y=\frac{x^2+2}{x+1}\)

=>y'=\(\frac{\left(x^2+2\right)^{\prime}\left(x+1\right)-\left(x^2+2\right)\left(x+1\right)^{\prime}}{\left(x+1\right)^2}=\frac{2x\left(x+1\right)-\left(x^2+2\right)}{\left(x+1\right)^2}=\frac{2x^2+2x-x^2-2}{\left(x+1\right)^2}=\frac{x^2+2x-2}{\left(x+1\right)^2}\)

Đặt y'>0

=>\(x^2+2x-2>0\)

=>\(x^2+2x+1>3\)

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

=>\(\left[\begin{array}{l}x+1>\sqrt3\\ x+1<-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\sqrt3-1\\ x<-\sqrt3-1\end{array}\right.\)

Vậy: Hàm số đồng biến trên các khoảng \(\left(\sqrt3-1;+\infty\right);\left(-\infty;-\sqrt3-1\right)\)

Đặt y'<0

=>\(x^2+2x-2<0\)

=>\(x^2+2x+1<3\)

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

=>\(-\sqrt3

=>\(-\sqrt3-1

=>Hàm số nghịch biến trên khoảng \(\left(-\sqrt3-1;\sqrt3-1\right)\)

b: \(y=\frac{2x^2-3}{x-2}\)

=>y'=\(\frac{\left(2x^2-3\right)^{\prime}\cdot\left(x-2\right)-\left(2x^2-3\right)\cdot\left(x-2\right)^{\prime}}{\left(x-2\right)^2}=\frac{2\cdot2x\left(x-2\right)-\left(2x^2-3\right)}{\left(x-2\right)^2}=\frac{4x^2-8x-2x^2+3}{\left(x-2\right)^2}=\frac{2x^2-8x+3}{\left(x-2\right)^2}\)

Đặt y'>0

=>\(2x^2-8x+3>0\)

=>\(x^2-4x+1,5>0\)

=>\(x^2-4x+4>\frac52\)

=>\(\left(x-2\right)^2>\frac52=\frac{10}{4}\)

=>\(\left[\begin{array}{l}x-2>\frac{\sqrt{10}}{2}\\ x-2<-\frac{\sqrt{10}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{4+\sqrt{10}}{2}\\ x<\frac{4-\sqrt{10}}{2}\end{array}\right.\)

=>Hàm số đồng biến trên các khoảng \(\left(\frac{4+\sqrt{10}}{2};+\infty\right);\left(-\infty;\frac{4-\sqrt{10}}{2}\right)\)

Đặt y'<0

=>\(2x^2-8x+3<0\)

=>\(x^2-4x+1,5<0\)

=>\(x^2-4x+4<\frac52\)

=>\(\left(x-2\right)^2<\frac52=\frac{10}{4}\)

=>\(-\frac{\sqrt{10}}{2}

=>\(\frac{4-\sqrt{10}}{2}

=>Hàm số nghịch biến trên khoảng \(\left(\frac{4-\sqrt{10}}{2};\frac{4+\sqrt{10}}{2}\right)\)

c:

ĐKXĐ: x<>2; x<>-2

\(y=\frac{x+1}{x^2-4}\)

=>y'=\(\frac{\left(x+1\right)^{\prime}\cdot\left(x^2-4\right)-\left(x+1\right)\left(x^2-4\right)^{\prime}}{\left(x^2-4\right)^2}=\frac{x^2-4-2x\left(x+1\right)}{\left(x^2-4\right)^2}\)

\(=\frac{x^2-4-2x^2-2x}{\left(x^2-4\right)^2}=\frac{-x^2-2x-4}{\left(x^2-4\right)^2}=\frac{-x^2-2x-1-3}{\left(x^2-4\right)^2}=\frac{-\left(x+1\right)^2-3}{\left(x^2-4\right)^2}<0\forall x\) thỏa mãn ĐKXĐ

=>Hàm số luôn nghịch biến trên mọi khoảng ở TXĐ
d: ĐKXĐ: x<>1; x<>-1

\(y=\frac{2x+3}{x^2-1}\)

=>y'=\(\frac{\left(2x+3\right)^{\prime}\cdot\left(x^2-1\right)-\left(2x+3\right)\left(x^2-1\right)^{\prime}}{\left(x^2-1\right)^2}\)

\(=\frac{2\left(x^2-1\right)-2x\left(2x+3\right)}{\left(x^2-1\right)^2}=\frac{2x^2-2-4x^2-6x}{\left(x^2-1\right)^2}=\frac{-2x^2-6x-2}{\left(x^2-1\right)^2}\)

\(=\frac{-2\left(x^2+3x+1\right)}{\left(x^2-1\right)^2}\)

Đặt y'<0

=>\(-2\left(x^2+3x+1\right)<0\)

=>\(x^2+3x+1>0\)

=>\(x^2+3x+\frac94>\frac54\)

=>\(\left(x+\frac32\right)^2>\frac54\)

=>\(\left[\begin{array}{l}x+\frac32>\frac{\sqrt5}{2}\\ x+\frac32<-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}x>\frac{\sqrt5-3}{2}\\ x<\frac{-\sqrt5-3}{2}\end{array}\right.\)

=>Hàm số nghịch biến trên các khoảng \(\left(\frac{\sqrt5-3}{2};+\infty\right);\left(-\infty;\frac{-\sqrt5-3}{2}\right)\)

Đặt y'>0

=>\(-2\left(x^2+3x+1\right)>0\)

=>\(x^2+3x+1<0\)

=>\(x^2+3x+\frac94<\frac54\)

=>\(\left(x+\frac32\right)^2<\frac54\)

=>\(-\frac{\sqrt5}{2}

=>\(\frac{-\sqrt5-3}{2}

=>Hàm số đồng biến trên khoảng \(\left(\frac{-\sqrt5-3}{2};\frac{\sqrt5-3}{2}\right)\)

a)

\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);                                                   

b)

\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);

c)

\(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) \\= 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 \\= 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);                                                               

d)

\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).

Lời giải

a)

\(\left(\frac{3}{2x-y}-\frac{2}{2x+y}-\frac{1}{2x-5y}\right).\frac{4x^2-y^2}{y^2}\)

\(=\frac{3(4x^2-y^2)}{(2x-y)y^2}-\frac{2(4x^2-y^2)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)

\(=\frac{3(2x-y)(2x+y)}{(2x-y)y^2}-\frac{2(2x-y)(2x+y)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)

\(=\frac{3(2x+y)-2(2x-y)}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)

\(=\frac{2x+5y}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)

\(=\frac{(2x+5y)(2x-5y)-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)

\(=\frac{4x^2-25y^2-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}=\frac{-25}{2x-5y}+\frac{1}{2x-5y}=\frac{-24}{2x-5y}\)

Ta có đpcm.

b) 

\(\frac{x^2-x+1}{x^2+x}.\frac{x+1}{3x-2}.\frac{9x-6}{x^2-x+1}\)

\(=\frac{(x^2-x+1)(x+1).3(3x-2)}{x(x+1)(3x-2)(x^2-x+1)}\)

\(=\frac{3}{x}\) (đpcm)

Cám ơn ạ :)

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