1/ ĐKXĐ: ...

\(\Leftrightarrow x=2016-2015\sqrt{x}-x\)

\(\Leftrightarrow2x+2015\sqrt{x}-2016=0\)

Đặt \(\sqrt{x}=t\ge0\)

\(\Rightarrow2t^2+2015t-2016=0\)

Nghiệm xấu kinh khủng, bạn tự giải

2. ĐKXĐ: ...

\(x^2+4x+4+4y^2-8y+4=4xy+13\)

\(\Leftrightarrow\left(x-2y\right)^2+4\left(x-2y\right)-5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2y=1\\x-2y=-5< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=2y+1\)

Thay xuống dưới:

\(\sqrt{\frac{\left(x+y\right)\left(x-2y\right)}{x-y}}+\sqrt{x+y}=\frac{2}{\sqrt{\left(x-y\right)\left(x+y\right)}}\)

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

\(\Leftrightarrow3y+1+\left(3y+1\right)\sqrt{y+1}=2\)

\(\Leftrightarrow6y+\left(3y+1\right)\left(\sqrt{y+1}-1\right)=0\)

\(\Leftrightarrow6y+\frac{\left(3y+1\right)y}{\sqrt{y+1}+1}=0\)

\(\Leftrightarrow y\left(6+\frac{3y+1}{\sqrt{y+1}+1}\right)=0\Rightarrow y=0\Rightarrow x=1\)

a/ Đơn giản là dùng phép thế:

\(x+2y+x+y+z=0\Rightarrow x+2y=0\Rightarrow x=-2y\)

\(x+y+z=0\Rightarrow z=-\left(x+y\right)=-\left(-2y+y\right)=y\)

Thế vào pt cuối:

\(\left(1-2y\right)^2+\left(y+2\right)^2+\left(y+3\right)^2=26\)

Vậy là xong

b/ Sử dụng hệ số bất định:

\(\left\{{}\begin{matrix}a\left(\frac{x}{3}+\frac{y}{12}-\frac{z}{4}\right)=a\\b\left(\frac{x}{10}+\frac{y}{5}+\frac{z}{3}\right)=b\end{matrix}\right.\)

\(\Rightarrow\left(\frac{a}{3}+\frac{b}{10}\right)x+\left(\frac{a}{12}+\frac{b}{5}\right)y+\left(\frac{-a}{4}+\frac{b}{3}\right)z=a+b\) (1)

Ta cần a;b sao cho \(\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}=-\frac{a}{4}+\frac{b}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{a}{3}+\frac{b}{10}=\frac{a}{12}+\frac{b}{5}\\\frac{a}{3}+\frac{b}{10}=-\frac{a}{4}+\frac{b}{3}\end{matrix}\right.\) \(\Rightarrow\frac{a}{2}=\frac{b}{5}\)

Chọn \(\left\{{}\begin{matrix}a=2\\b=5\end{matrix}\right.\) thay vào (1):

\(\frac{7}{6}\left(x+y+z\right)=7\Rightarrow x+y+z=6\)

\(e,\left\{{}\begin{matrix}\left(\frac{x}{y}\right)^3+\left(\frac{x}{y}\right)^2=12\\\left(xy\right)^2+xy=6\end{matrix}\right.\left(x;y\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy\in\left\{2;-3\right\}\end{matrix}\right.\)

Vì \(\frac{x}{y}=2>0\Rightarrow xy>0\Rightarrow xy=2\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{y}=2\\xy=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2y\\2y^2=2\end{matrix}\right.\)

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

\(a,\left\{{}\begin{matrix}x^2+\frac{1}{y^2}+\frac{x}{y}=3\\x+\frac{1}{y}+\frac{x}{y}=3\end{matrix}\right.\left(x;y\ne0\right)\)

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

Đặt \(\left\{{}\begin{matrix}x+\frac{1}{y}=a\\\frac{x}{y}=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a^2-b=3\\a+b=3\end{matrix}\right.\)

Làm nốt nha

\(\left\{{}\begin{matrix}\sqrt{x-1}+\sqrt{y-1}=2\\\frac{1}{x}+\frac{1}{y}=1\end{matrix}\right.\left(x;y\ge1\right)\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-\left(x+y\right)+1}=6\\x+y=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+2\sqrt{xy-xy+1}=6\\x+y=xy\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=4\end{matrix}\right.\)

Làm nốt

e) Sửa đề: \(\left\{{}\begin{matrix}x\left(x^2-y^2\right)+x^2=2\sqrt{\left(x-y^2\right)^3}\\76x^2-20y^2+2=\sqrt[3]{4x\left(8x+1\right)}\end{matrix}\right.\)

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

Đặt \(\sqrt{x-y^2}=a.\text{Thay vào, ta có: }x^3+xa^2-2a^3=0\)

Làm tiếp như ở Câu hỏi của Nguyễn Mai - Toán lớp 9 - Học toán với OnlineMath

Băng Băng 2k6, Vũ Minh Tuấn, Nguyễn Việt Lâm, HISINOMA KINIMADO, Akai Haruma, Inosuke Hashibira, Nguyễn Thị Ngọc Thơ, Nguyễn Lê Phước Thịnh, Quân Tạ Minh, An Võ (leo), @tth_new

e nhiều bài quá giải k kịp mn giúp e vs ạ!cần gấp lắm ạ

thanks nhiều!

a/ \(\left\{{}\begin{matrix}\left(x^2+x\right)+\left(y^2+y\right)=18\\\left(x^2+x\right)\left(y^2+y\right)=72\end{matrix}\right.\)

Theo Viet đảo, \(x^2+x\) và \(y^2+y\) là nghiệm của:

\(t^2-18t+72=0\Rightarrow\left[{}\begin{matrix}t=12\\t=6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2+x=6\\y^2+y=12\end{matrix}\right.\\\left\{{}\begin{matrix}x^2+x=12\\y^2+y=6\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\left\{2;-3\right\}\\y=\left\{3;-4\right\}\end{matrix}\right.\\\left\{{}\begin{matrix}x=\left\{3;-4\right\}\\y=\left\{2;-3\right\}\end{matrix}\right.\end{matrix}\right.\)

b/ ĐKXĐ: ...

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

Nhận thấy \(y=\frac{1}{3}\) không phải nghiệm

\(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x}+\frac{1}{y+1}=1\\\frac{1}{x}=\frac{y}{3y-1}\end{matrix}\right.\) \(\Rightarrow\frac{y}{3y-1}+\frac{1}{y+1}=1\)

\(\Leftrightarrow y\left(y+1\right)+3y-1=\left(3y-1\right)\left(y+1\right)\)

\(\Leftrightarrow y^2-y=0\Rightarrow\left[{}\begin{matrix}y=0\left(l\right)\\y=1\end{matrix}\right.\) \(\Rightarrow x=2\)

c/ ĐKXĐ: ...

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

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

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

\(\Rightarrow y+1=\left(3-y\right)y\)

\(\Leftrightarrow y^2-2y+1=0\Rightarrow y=1\Rightarrow x=2\)