Ta có : 3x² - 7xy + 4y² = 0

=> 3x² - 3xy - 4xy + 4y² = 0

=> 3x( x - y) - 4y( x - y) = 0

=> ( x - y)( 3x - 4y) = 0

=> x = y ; 3x = 4y

Thay : x = y ; 3x = 4y vào phân thức trên ta có:

\(A=\dfrac{4y+2x}{5y-7x}+\dfrac{3x-2y}{10y-4x}\)

\(A=\dfrac{3x+2x}{5x-7x}+\dfrac{4y-2y}{10x-4x}\)

\(A=\dfrac{5x}{-2x}+\dfrac{2y}{6x}=\dfrac{5}{-2}+\dfrac{1}{3}=\dfrac{-13}{6}\)

\(\dfrac{15x^3y^5-10x^4y^4-21x^5y^3z}{-6x^3y^2}=-\dfrac{5}{2}y^3+\dfrac{5}{3}xy^2+\dfrac{7}{2}x^2yz\)

\(x=-\dfrac{1}{2}=-0.5,y=\dfrac{5}{4}=1.25\\x=2,y=\dfrac{7}{2}=3.5\)

a, (3x-1)(x²+2)=(3x-1)(7x-10)

<=>(3x-1)(x²+2)-(3x-1)(7x-10)=0

<=>(3x-1)(x²+2-7x+10)=0

<=>(3x-1)(x²-7x+12)=0

<=>(3x-1)(x²-3x-4x+12)=0

<=>(3x-1)(x-3)(x-4)=0

<=>\(\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\)<=>\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy ft có tập nghiệm S=\(\left\{\dfrac{1}{3},3,4\right\}\)

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (ĐKXĐ:t\(\ne2;t\ne-3\))

<=>\(\dfrac{\left(t+3\right)^2+\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{t^2-2t+3t-6}\)

<=>\(\dfrac{t^2+6t+9+t^2-4t+4}{\left(t-2\right)\left(t+3\right)}\)=\(\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

=>2t²+2t+13=5t+15

<=>2t²+2t-5t+13-15=0

<=>2t²-3t-2=0

<=>2t²-4t+t-2=0

<=>(t-2)(2t+1)=0

<=>\(\left[{}\begin{matrix}t-2=0\\2t+1=0\end{matrix}\right.< =>\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-1}{2}\left(tmđkxđ\right)\end{matrix}\right.\)

Vậy ft có nghiệm duy nhất x=\(\dfrac{-1}{2}\)

Giải:

a) \(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

Chia cả hai vế cho 3x-1, ta được:

\(x^2+2=7x-10\)

\(\Leftrightarrow x^2-7x+10+2=0\)

\(\Leftrightarrow x^2-7x+12=0\)

\(\Leftrightarrow x^2-4x-3x+12=0\)

\(\Leftrightarrow x\left(x-4\right)-3\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

Vậy ...

b) \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) (1)

ĐKXĐ: \(t\ne2;t\ne-3\)

\(\left(1\right)\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t-2\right)\left(t+3\right)}\)

\(\Rightarrow\left(t+3\right)^2+\left(t-2\right)^2=5t+15\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)

\(\Leftrightarrow2t^2+2t+13=5t+15\)

\(\Leftrightarrow2t^2+2t+13-5t-15=0\)

\(\Leftrightarrow2t^2-3t-2=0\)

\(\Leftrightarrow2t^2-4t+t-2=0\)

\(\Leftrightarrow2t\left(t-2\right)+\left(t-2\right)=0\)

\(\Leftrightarrow\left(2t+1\right)\left(t-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\left(tm\right)\\t=2\left(ktm\right)\end{matrix}\right.\)

Vậy ...

a,\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x^2-7x+12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=1\\\left(x-4\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x-4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\\x=3\end{matrix}\right.\)

Vậy...

b,\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)

\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Leftrightarrow\dfrac{\left(t+3\right)^2}{\left(t-2\right)\left(t+3\right)}+\dfrac{\left(t-2\right)^2}{\left(t-2\right)\left(t+3\right)}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4=5t+15\)

\(\Leftrightarrow t^2+t^2+6t-4t-5t=15-9-4\)

\(\Leftrightarrow2t^2-3t=2\)

\(\Leftrightarrow\left(2t+1\right)\left(t+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2t+1=0\\t+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{1}{2}\\t=-2\end{matrix}\right.\)

Vậy...

\(=2y^2+3xy-x^2\)

Lời giải:

\(A=x^2-5x+5\)

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

\(=\left(x-\frac{5}{2}\right)^2-\frac{5}{4}\)

Vì \((x-\frac{5}{2})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow A=(x-\frac{5}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}\)

Vậy \(A_{\min}=-\frac{5}{4}\). Dấu bằng xảy ra khi \(x=\frac{5}{2}\)

--------------

\(B=x^2-3x+1\)

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

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

Vì \((x-\frac{3}{2})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow B\geq 0-\frac{5}{4}=-\frac{5}{4}\)

Vậy \(B_{\min}=\frac{-5}{4}\Leftrightarrow x=\frac{3}{2}\)

\(C=3x^2-6x+8\)

\(=3(x^2-2x+1)+5\)

\(=3(x-1)^2+5\)

Vì \((x-1)^2\geq 0, \forall x\in\mathbb{R}\Rightarrow C\geq 3.0+5=5\)

Do đó \(C_{\min}=5\Leftrightarrow x=1\)

----------------

\(D=7x^2+21x+3\)

\(=7[x^2+3x+(\frac{3}{2})^2]-\frac{51}{4}\)

\(=7[x^2+2.\frac{3}{2}.x+(\frac{3}{2})^2]-\frac{51}{4}=7(x+\frac{3}{2})^2-\frac{51}{4}\)

Vì \((x+\frac{3}{2})^2\geq 0, \forall x\in\mathbb{R}\Rightarrow D\geq 7.0-\frac{51}{4}=\frac{-51}{4}\)

Vậy \(D_{\min}=-\frac{51}{4}\Leftrightarrow x=\frac{-3}{2}\)

\(E=x^2+y^2+2x+4y\)

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

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

\(=(x+1)^2+(y+2)^2-5\)

Vì \((x+1)^2; (y+2)^2\geq 0, \forall x,y\in\mathbb{R}\), do đó:

\(E\geq 0+0-5=-5\)

Vậy \(E_{\min}=-5\).

Dấu bằng xảy ra khi \((x+1)^2=(y+2)^2=0\Leftrightarrow x=-1; y=-2\)

a) ĐK: \(x^2+7x+7\ge0\)

Đặt \(a=\sqrt{x^2+7x+7}\)  \(\left(a\ge0\right)\)

PT \(\Rightarrow3a^2-3+2a=2\) \(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2+7x+7=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)  (Thỏa mãn) 

Vậy ...

b) ĐK: \(x^2-6x+6\ge0\)

Đặt \(a=\sqrt{x^2-6x+6}\)  \(\left(a\ge0\right)\)

PT \(\Rightarrow a^2+3=4a\) \(\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)  (Thỏa mãn)

+) Với \(a=3\) \(\Rightarrow x^2-6x+6=9\) \(\Leftrightarrow\left[{}\begin{matrix}x=3+2\sqrt{3}\\x=3-2\sqrt{3}\end{matrix}\right.\)  (Thỏa mãn)

+) Với \(a=1\) \(\Rightarrow x^2-6x+6=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)  (Thỏa mãn)

  Vậy ...

c)C1: Áp dụng bđt AM-GM \(\Rightarrow VT\ge2>\dfrac{7}{4}\)

=> Dấu = ko xảy ra hay pt vô nghiệm

C2: Đk:\(x>0\)

Đặt \(a=\sqrt{\dfrac{x^2+x+1}{x}}\left(a>0\right)\) \(\Rightarrow\dfrac{1}{a}=\sqrt{\dfrac{x}{x^2+x+1}}\)

Pttt: \(a+\dfrac{1}{a}=\dfrac{7}{4}\Leftrightarrow4a^2-7a+4=0\) 

\(\Delta =-15<0 \) => Pt vô nghiệm

Vậy...

d) Đk: \(x\le-8;x\ge0\)

Đặt \(t=\sqrt{x\left(8+x\right)}\left(t\ge0\right)\)

Pttt: \(t^2-3=2t\Leftrightarrow t^2-2t-3=0\Leftrightarrow\left[{}\begin{matrix}t=3\left(tm\right)\\t=-1\left(ktm\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x\left(8+x\right)}=3\Leftrightarrow x^2+8x-9=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\) (tm)

Vậy...

a: \(=\dfrac{x+3}{\left(x-1\right)\left(x+1\right)}-\dfrac{1}{x\left(x+1\right)}\)

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

b: \(=\dfrac{24y^5}{7x^2}\cdot\dfrac{-21x}{12y^3}=2y^2\cdot\dfrac{-3}{x}=\dfrac{-6y^2}{x}\)

c: \(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x-1\right)\left(x+1\right)}=\dfrac{-1}{2\left(x+1\right)}\)

d: \(=\dfrac{7x+2}{3\left(2x-y\right)}\cdot\dfrac{6x\left(2x-y\right)}{2\left(7x+2\right)}=x\)

1 x mũ 2 + 4xy + 4y mũ 2 = x^2 + 4xy + 4y^2 =(2y+x)^2

2,       4x mũ 2 - 36y mũ 2 =4x^2 -36y^2 = -4 (3y-x) (3y+x)

1, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)

2, \(4x^2-36y^2=\left(2x-6y\right)\left(2x+6y\right)\)

3, \(x^3-2x^2+5x-10=x^2\left(x-2\right)+5\left(x-2\right)=\left(x^2+5\right)\left(x-2\right)\)

4, \(a^3+a^2+3a+3=a^2\left(a+1\right)+3\left(a+1\right)=\left(a^2+3\right)\left(a+1\right)\)

5, \(7x^3-21x^2+3-x=7x^2\left(x-3\right)-\left(x-3\right)=\left(7x^2-1\right)\left(x-3\right)\)