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

\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}\\6x+2y=-2\end{matrix}\right.-6x+12y=18}\)

a.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge2\\y\ge3\end{matrix}\right.\)

\(\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\5\sqrt{x-2}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3\sqrt{x-2}+3\sqrt{y-3}=9\\\sqrt{x-2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2}=1\\\sqrt{y-3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)

b.

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{4x}{x+1}-\dfrac{10}{y+4}=8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{15x}{x+1}+\dfrac{10}{y+4}=20\\\dfrac{19x}{x+1}=28\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{x+1}=\dfrac{28}{19}\\\dfrac{1}{y+4}=-\dfrac{4}{19}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=28x+28\\4y+16=-19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{28}{9}\\y=-\dfrac{35}{4}\end{matrix}\right.\)

Xem lại giúp tớ dấu căn ở câu c và d nhé.  

a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

Câu 4:

Giả sử điều cần chứng minh là đúng

\(\Rightarrow x=y\), thay vào điều kiện ở đề bài, ta được:

\(\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}=\sqrt{x+2014}+\sqrt{2015-x}-\sqrt{2014-x}\) (luôn đúng)

Vậy điều cần chứng minh là đúng

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

⇔ \(\sqrt{\left(x-4\right)\left(x-1\right)}-2\sqrt{x-4}+2\sqrt{x+5}-\sqrt{\left(x+5\right)\left(x-1\right)}=0\)

⇔ \(\sqrt{x-4}.\left(\sqrt{x-1}-2\right)-\sqrt{x+5}\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left(\sqrt{x-4}-\sqrt{x+5}\right)\left(\sqrt{x-1}-2\right)=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}-\sqrt{x+5}=0\\\sqrt{x-1}-2=0\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-4}=\sqrt{x+5}\\\sqrt{x-1}=2\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}x\in\varnothing\\x=5\end{matrix}\right.\)

⇔ x = 5

Vậy S = {5}

Bài 1:

ĐKĐB suy ra $x(x+1)+y(y+1)=3x^2+xy-4x+2y+2$

$\Leftrightarrow 2x^2+x(y-5)+(y-y^2+2)=0$

Coi đây là PT bậc 2 ẩn $x$

$\Delta=(y-5)^2-4(y-y^2+2)=(3y-3)^2$Do đó:

$x=\frac{y+1}{2}$ hoặc $x=2-y$. Thay vào một trong 2 phương trình ban đầu ta thu được:

$(x,y)=(\frac{-4}{5}, \frac{-13}{5}); (1,1)$

2) Ta có: \(\left\{{}\begin{matrix}\sqrt{3x-1}-\sqrt{2y+1}=1\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3x-1}-2\sqrt{2y+1}=2\\2\sqrt{3x-1}+3\sqrt{2y+1}=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-5\sqrt{2y+1}=-10\\\sqrt{3x-1}-\sqrt{2y+1}=1\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2y+1=4\\3x-1=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2y=3\\3x=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{3}{2}\\x=\dfrac{10}{3}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{10}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-2}+2\sqrt{y-3}=6\\2\sqrt{x-2}-3\sqrt{y-3}=-4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y-3}=10\\\sqrt{x-2}+\sqrt{y-3}=3\end{matrix}\right.\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}y-3=4\\x-2=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=7\\x=3\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=3\\y=7\end{matrix}\right.\)

bài này mình chưa giải dc triệt để ở cái cuối

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

\(\Leftrightarrow4x^3-8x^2+6x-2=2x^3\left(4-2y\right)\sqrt{3-2y}\left(1\right)\)

\(đặt:\sqrt{3-2y}=a\ge0\Rightarrow a^2+1=4-2y\)

\(\left(1\right)\Leftrightarrow4x^3-8x^2+6x-2=2x^3.\left(a^2+1\right)a\)

\(\Leftrightarrow4x^3-8x^2+6x-2-2x^3\left(a^2+1\right)a\)

\(\Leftrightarrow-2\left(xa-x+1\right)\left[\left(xa\right)^2+x^2a+2x^2-xa-2x+1\right]=0\)

\(\Leftrightarrow x.a-x+1=0\Leftrightarrow x\left(a-1\right)=-1\Leftrightarrow x=\dfrac{-1}{a-1}\)

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

\(=\dfrac{-a}{a-1}-\sqrt{\dfrac{-1}{a-1}}\)

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

\(\Rightarrow\left(\dfrac{-a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\left(\sqrt{\dfrac{-a}{a-1}+2}+\sqrt{\dfrac{a-2}{a-1}}\right)-4=0\)

\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right).2\sqrt{\dfrac{a-2}{a-1}}=4\)

\(\Leftrightarrow\left(-\dfrac{a}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{\dfrac{a-2}{a-1}}=2\)

\(\Leftrightarrow\left(-1+\dfrac{-1}{a-1}-\sqrt{-\dfrac{1}{a-1}}\right)\sqrt{1-\dfrac{1}{a-1}}=2\)(3)

\(đặt:1-\dfrac{1}{a-1}=u\Rightarrow\sqrt{-\dfrac{1}{a-1}}=\sqrt{u-1}\)

\(\left(3\right)\Leftrightarrow\left(u-2-\sqrt{u-1}\right)\sqrt{u}=2\)

bình phương lên tính được u

\(\Rightarrow u=.....\Rightarrow a\Rightarrow y=...\Rightarrow x=....\)

Với \(x=0\) không phải nghiệm

Với \(x>0\) chia 2 vế cho pt đầu cho \(x^3\)

\(\Rightarrow2-\dfrac{4}{x}+\dfrac{3}{x^2}-\dfrac{1}{x^3}=2\left(2-y\right)\sqrt{3-2y}\)

\(\Leftrightarrow1-\dfrac{1}{x}+\left(1-\dfrac{1}{x}\right)^3=\sqrt{3-2y}+\sqrt{\left(3-2y\right)^3}\)

Xét hàm \(f\left(t\right)=t+t^3\Rightarrow f'\left(t\right)=1+3t^2>0\Rightarrow f\left(t\right)\) đồng biến

\(\Rightarrow1-\dfrac{1}{x}=\sqrt{3-2y}\)

Thế vào pt dưới:

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

\(\Leftrightarrow\left(x-\sqrt{x}-1\right)\left(\sqrt{x+1}+\sqrt{x+1}\right)=4\)

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

Phương trình này ko có nghiệm đẹp, chắc bạn ghi nhầm đề bài của pt dưới

... giải ra \(1-\dfrac{1}{x}=\sqrt{3-2y}\)

Thế xuống pt dưới:

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

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

Có vẻ đề bài vẫn sai

Do \(x\ge1\) theo ĐKXĐ nên \(x+1\ge2\) ; \(\left(\sqrt{x+1}+\sqrt{x-1}\right)^4\ge\left(\sqrt{2}+0\right)^4=4\)

\(\Rightarrow\left(x+1\right)\left(\sqrt{x+1}+\sqrt{x-1}\right)^4\ge8>4\) nên pt vô nghiệm