a.Hệ thứ nhất kì quặc thật:

\(\Leftrightarrow\sqrt{y^2+xy}+\sqrt{x+y}=\sqrt{x^2+y^2}+2\)

\(\Leftrightarrow\sqrt{x^2+y^2}-\sqrt{y^2+xy}=\sqrt{x+y}-2\)

\(\Leftrightarrow\dfrac{x\left(x-y\right)}{\sqrt{x^2+y^2}+\sqrt{y^2+xy}}=\dfrac{x+y-4}{\sqrt{x+y}+2}\)

\(\Rightarrow\left(x-y\right)\left(x+y-4\right)=\left(\dfrac{\sqrt{x^2+y^2}+\sqrt{y^2+xy}}{x\sqrt{x+y}+2x}\right)\left(x+y-4\right)^2\ge0\) (1)

\(2.\dfrac{x}{2}\sqrt{y-1}+2.\dfrac{y}{2}\sqrt{x-1}\le\dfrac{x^2}{4}+y-1+\dfrac{y^2}{4}+x-1\)

\(\Rightarrow\dfrac{x^2+4y-4}{2}\le\dfrac{x^2+y^2+4x+4y-8}{4}\)

\(\Leftrightarrow x^2-y^2+4y-4x\le0\)

\(\Leftrightarrow\left(x-y\right)\left(x+y-4\right)\le0\) (2)

(1);(2) \(\Rightarrow\left(x-y\right)\left(x+y-4\right)=0\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=2\)

b.

\(x^3-x^2y+2y^2-2xy=0\)

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

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

\(\Leftrightarrow y=x\) (loại \(x^2-2y=0\) do ĐKXĐ \(x^2-2y-1\ge0\))

Thế vào pt dưới

\(2\sqrt{x^2-2x-1}+\sqrt[3]{x^3-14}=x-2\)

\(\Leftrightarrow2\sqrt{x^2-2x-1}+\dfrac{x^3-14-\left(x-2\right)^3}{\sqrt[3]{\left(x^3-14\right)^2}+\left(x-2\right)\sqrt[3]{x^3-14}+\left(x-2\right)^2}=0\)

\(\Leftrightarrow\sqrt[]{x^2-2x-1}\left(2+\dfrac{6\sqrt[]{x^2-2x-1}}{\sqrt[3]{\left(x^3-14\right)^2}+\left(x-2\right)\sqrt[3]{x^3-14}+\left(x-2\right)^2}\right)=0\)

\(\Leftrightarrow\sqrt{x^2-2x-1}=0\)

2)ĐK:x\(\ge\frac{1}{2}\)

pt(2)\(\Leftrightarrow\left(y+1\right)^3\)+(y+1)=\(\left(2x\right)^3\)+2x

Xét hàm số: f(t)=\(t^3\)+t

f'(t)=3\(t^2\)+1>0,\(\forall\)t

\(\Rightarrow\)hàm số liên tục và đồng biến trên R

\(\Rightarrow\)y+1=2x

Thay y=2x-1 vào pt(1) ta đc:

\(x^2\)-2x=2\(\sqrt{2x-1}\)

\(\Leftrightarrow\left(x^2-4x+2\right)\left(1+\frac{4}{2x-2+2\sqrt{2x-1}}\right)=0\)

\(\Leftrightarrow x^2\)-4x+2=0(do(...)>0)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2+\sqrt{2}\Rightarrow y=3+2\sqrt{2}\\x=2-\sqrt{2}\Rightarrow y=3-2\sqrt{2}\end{array}\right.\)

4)ĐK:\(y\ge\frac{2}{3}\)

pt(1)\(\Leftrightarrow x-\sqrt{3y-2}=\sqrt{3y\left(3y-2\right)}-x\sqrt{x^2+2}\)

\(\Leftrightarrow x\left(\sqrt{x^2+2}+1\right)=\sqrt{3y-2}\left(\sqrt{3y}+1\right)\)

Xét hàm số:\(f\left(t\right)=t\left(\sqrt{t^2+2}+1\right)\)

\(\Rightarrow\)hàm số liên tục và đồng biến trên R

\(\Rightarrow x=\sqrt{3y-2}\)

Thay vào pt(2) ta đc:\(\sqrt{3y-2}+y+\sqrt{y+3}=4\)

\(\Leftrightarrow\sqrt{3y-2}-1+\sqrt{y+3}-2+y-1=0\)

\(\Leftrightarrow\left(y-1\right)\left(\frac{3}{\sqrt{3y-2}+1}+\frac{1}{\sqrt{y+3}+2}+1\right)=0\)

\(\Leftrightarrow y=1\Rightarrow x=1\)(do...)>0)

KL:...

a/ \(y'=\dfrac{\left(x^3+2\sqrt{x-1}\right)'\left(x-1\right)-\left(x-1\right)'\left(x^3+2\sqrt{x-1}\right)}{\left(x-1\right)^2}\)

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

b/ \(y'=\dfrac{\left(4x^3+2x-3\right)'\left(\sqrt{x^2+2}\right)-\left(\sqrt{x^2+2}\right)'\left(4x^3+2x-3\right)}{x^2+2}\)

\(y'=\dfrac{\left(12x^2+2\right)\sqrt{x^2+2}-\dfrac{x}{\sqrt{x^2+2}}\left(4x^3+2x-3\right)}{x^2+2}\) (ban tu rut gon nhe)

c/ \(y'=\dfrac{\left(x^3+x+1\right)'\left(x^3+x+1\right)}{\left|x^3+x+1\right|}=\dfrac{\left(3x^2+1\right)\left(x^3+x+1\right)}{\left|x^3+x+1\right|}\) 

d/ \(y'=\dfrac{3x^2-24x^3}{2\sqrt{x^3-6x^4+7}}\)

e/ \(y'=\dfrac{\left(x^5+1\right)'\left(2-\sqrt{x^2+3}\right)-\left(x^5+1\right)\left(2-\sqrt{x^2+3}\right)'}{\left(2-\sqrt{x^2+3}\right)^2}\)

\(y'=\dfrac{5x^4\left(2-\sqrt{x^2+3}\right)+\left(x^5+1\right)\dfrac{x}{\sqrt{x^2+3}}}{\left(2-\sqrt{x^2+3}\right)^2}\)

a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)

a.

ĐKXĐ: $x\geq 0; y\geq 1$

PT $\Leftrightarrow (x-4\sqrt{x}+4)+(y-1-6\sqrt{y-1}+9)=0$
$\Leftrightarrow (\sqrt{x}-2)^2+(\sqrt{y-1}-3)^2=0$
Vì $(\sqrt{x}-2)^2; (\sqrt{y-1}-3)^2\geq 0$ với mọi $x\geq 0; y\geq 1$ nên để tổng của chúng bằng $0$ thì:

$\sqrt{x}-2=\sqrt{y-1}-3=0$

$\Leftrightarrow x=4; y=10$

b.

ĐKXĐ: $x\geq -1; y\geq -2; z\geq -3$
PT $\Leftrightarrow x+y+z+35-4\sqrt{x+1}-6\sqrt{y+2}-8\sqrt{z+3}=0$

$\Leftrightarrow [(x+1)-4\sqrt{x+1}+4]+[(y+2)-6\sqrt{y+2}+9]+[(z+3)-8\sqrt{z+3}+16]=0$

$\Leftrightarrow (\sqrt{x+1}-2)^2+(\sqrt{y+2}-3)^2+(\sqrt{z+3}-4)^2=0$
$\Rightarrow \sqrt{x+1}-2=\sqrt{y+2}-3=\sqrt{z+3}-4=0$
$\Rightarrow x=3; y=7; z=13$

c.

ĐKXĐ: $x\geq \frac{-1}{8}$

PT $\Leftrightarrow 9x+17-6\sqrt{8x+1}-4\sqrt{x+3}=0$

$\Leftrightarrow [(8x+1)-6\sqrt{8x+1}+9]+[(x+3)-4\sqrt{x+3}+4]=0$

$\Leftrightarrow (\sqrt{8x+1}-3)^2+(\sqrt{x+3}-2)^2=0$

$\Rightarrow \sqrt{8x+1}-3=\sqrt{x+3}-2=0$

$\Rightarrow x=1$ (thỏa mãn đkxđ)

1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)

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

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

4. Đk: \(x,y\ge0\)

\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>

Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)

*Công thức: Biến đổi x theo y và ngc lại và dùng các quy tắc.

a)\(\left\{{}\begin{matrix}\sqrt{2}x-\sqrt{3}y=1\\x+\sqrt{3}y=\sqrt{2}\left(1\right)\end{matrix}\right.\)

Cộng 2 pt ta đc: x=1

Thay vào (1):\(\Leftrightarrow y=\frac{\sqrt{2}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\)

Vậy (x;y)\(=\left(1;\frac{\sqrt{6}}{3}\right)\)

Những câu sau làm ttự.

#Walker