bạn ghi sai đề r

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/ \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)

Do \(x=y;x=-y\) đều ko phải nghiệm

\(\Rightarrow\frac{x^2+y^2}{\left(x+y\right)^2}=\frac{13}{25}\Leftrightarrow25\left(x^2+y^2\right)=13\left(x+y\right)^2\)

\(\Leftrightarrow12x^2-26xy+12y^2=0\)

\(\Leftrightarrow\left(2x-3y\right)\left(3x-2y\right)=0\Rightarrow\left[{}\begin{matrix}y=\frac{2}{3}x\\y=\frac{3}{2}x\end{matrix}\right.\)

Thay vào 1 trong 2 pt ban đầu là xong

b/ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\y\ge0\end{matrix}\right.\) \(\Rightarrow x+y>0\)

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

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

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

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

Thay vào pt dưới:

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

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

\(\Leftrightarrow y\left(9y^2+6y+1\right)=2\left(y^2+2y+1\right)\)

\(\Leftrightarrow9y^3+2y^2-3y-2=0\)

Nghiệm quá xấu, bạn coi lại đề

c/ \(y=0\) không phải nghiệm

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+1+y\left(x+y\right)=4y\\y\left(x+y\right)^2-2\left(x^2+1\right)=7y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\frac{x^2+1}{y}+x+y=4\\\left(x+y\right)^2-2\left(\frac{x^2+1}{y}\right)=7\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=a\\\frac{x^2+1}{y}=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=4\\a^2-2b=7\end{matrix}\right.\) \(\Rightarrow a^2-2\left(4-a\right)=7\)

\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\Rightarrow b=1\\a=-5\Rightarrow b=9\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+y=3\\\frac{x^2+1}{y}=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=3-x\\x^2+1-y=0\end{matrix}\right.\)

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

TH2: làm tương tự

1,ĐK: \(x,y\ne-2\)

HPT<=> \(\left\{{}\begin{matrix}x\left(x+2\right)+y\left(y+2\right)=\left(x+2\right)\left(y+2\right)\left(1\right)\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x^2\left(x+2\right)^2+2xy\left(x+2\right)\left(y+2\right)+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\\x^2\left(x+2\right)^2+y^2\left(y+2\right)^2=\left(x+2\right)^2\left(y+2\right)^2\end{matrix}\right.\)

=> \(2xy\left(x+2\right)\left(y+2\right)=0\)

<=>\(2xy=0\) (do x+2 và y+2 \(\ne0\))

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

Tại x=0 thay vào (1) có: \(y\left(y+2\right)=2\left(y+2\right)\) <=> y= \(\pm2\) => y=2 (vì y khác -2)

Tại y=0 thay vào (1) có: \(x\left(x+2\right)=2\left(x+2\right)\) => x=2

Vậy HPT có 2 nghiệm duy nhất (2,0),(0,2)

2, ĐK: \(y\ne-1\)

HPT <=> \(\left\{{}\begin{matrix}x^2=2\left(x+3\right)\left(y+1\right)\left(1\right)\\\frac{3x^2}{y+1}=4-x\end{matrix}\right.\)

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

<=> 6(x+3)=4-x

<=> \(14=-7x\)

<=> \(x=-2\) thay vào (1) có \(4=2\left(y+1\right)\)

<=>y=1\(\)( tm)

Vậy hpt có một nghiệm duy nhất (-2,1)

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

PT (1) <=> \(\left(x-y\right)\left(x+y\right)+\left(x-y\right)=0\)

<=> (x-y)(x+y+1)=0

<=>\(\left[{}\begin{matrix}x=y\\y=-x-1\end{matrix}\right.\)

Tại x=y thay vào (2) có \(y^2-y=y+3\) <=> \(y^2-2y-3=0\) <=> (y-3)(y+1)=0 <=> \(\left[{}\begin{matrix}y=3\\y=-1\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Tại y=-1-x thay vào (2) có: \(x^2-x=-1-x+3\) <=> \(x^2=2\) <=> \(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}y=-1-\sqrt{2}\\y=-1+\sqrt{2}\end{matrix}\right.\)

Vậy hpt có 4 nghiệm (3,3),(-1,-1), ( \(\sqrt{2},-1-\sqrt{2}\)),( \(-\sqrt{2},-1+\sqrt{2}\))

4,\(\left\{{}\begin{matrix}x+y+\frac{1}{x}+\frac{1}{y}=\frac{9}{2}\left(1\right)\\xy+\frac{1}{xy}+\frac{x}{y}+\frac{y}{x}=5\left(2\right)\end{matrix}\right.\)(đk:\(x\ne0,y\ne0\))

<=> \(\left\{{}\begin{matrix}\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)=\frac{9}{2}\\\left(y+\frac{1}{y}\right)\left(x+\frac{1}{x}\right)=5\end{matrix}\right.\)

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

Có \(\left\{{}\begin{matrix}u+v=\frac{9}{2}\\uv=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\v\left(\frac{9}{2}-v\right)=5\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left(v-\frac{5}{2}\right)\left(v-2\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}u=\frac{9}{2}-v\\\left[{}\begin{matrix}v=\frac{5}{2}\\v=2\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\\\left[{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\end{matrix}\right.\)

Tại \(\left\{{}\begin{matrix}v=\frac{5}{2}\\u=2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=2\\y+\frac{1}{y}=\frac{5}{2}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)\left(y-\frac{1}{2}\right)=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=1\\\left[{}\begin{matrix}y=2\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\y=2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\y=\frac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Tại \(\left\{{}\begin{matrix}v=2\\u=\frac{5}{2}\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x+\frac{1}{x}=\frac{5}{2}\\y+\frac{1}{y}=2\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\left(x-2\right)\left(x-\frac{1}{2}\right)=0\\\left(y-1\right)^2=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\\y=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\\left[{}\begin{matrix}x=\frac{1}{2}\\y=1\end{matrix}\right.\end{matrix}\right.\)

Vậy hpt có 4 nghiệm (1,2),( \(1,\frac{1}{2}\)) ,( 2,1),(\(\frac{1}{2},1\)).

10.

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

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

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

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

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

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

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

@Nguyễn Việt Lâm @Lê Thị Thục Hiền @Akai Haruma @Trần Thanh Phương

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-2x+2y^2-y-1=0\\2y^2+2x+y+1-6xy=0\end{matrix}\right.\)

Cộng vế với vế:

\(2x^2+4y^2-6xy=0\)

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

Thế vào 1 trong 2 pt ban đầu

a, \(\left\{{}\begin{matrix}\left(x-y\right)\left(x^2+y^2\right)=13\\\left(x+y\right)\left(x^2-y^2\right)=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-y\right)\left(x^2+y^2\right)=26\\\left(x-y\right)\left(x+y\right)^2=25\end{matrix}\right.\)

Trừ vế theo vế \(pt\left(1\right)\) cho \(pt\left(2\right)\) ta được:

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

\(\Leftrightarrow\left(x-y\right)^3=1\)

\(\Leftrightarrow x-y=1\)

Khi đó hệ trở thành:

\(\left\{{}\begin{matrix}x^2+y^2=13\\\left(x+y\right)^2=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\13+2xy=25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2=13\\2xy=12\end{matrix}\right.\)

Cộng vế theo vế 2 phương trình:

\(\left(x+y\right)^2=25\)

\(\Leftrightarrow x+y=\pm5\)

TH1: \(x+y=5\)

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

TH2: \(x+y=-5\)

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

b, \(\left\{{}\begin{matrix}2x^2+x-\dfrac{1}{y}=2\\y-y^2x-2y^2=-2\end{matrix}\right.\)

ĐK: \(y\ne0\)

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

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

Đặt \(\dfrac{1}{y}=t\), hệ trở thành:

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

\(\Rightarrow\left(x-t\right)\left(x+t+1\right)=0\)

\(\Leftrightarrow...\)