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1) Đk: x khác -3
x khác 1
Biểu thức \(\Leftrightarrow\dfrac{x^2-x}{x^2+2x-3}+\dfrac{2x+6}{x^2+2x-3}=\dfrac{12}{x^2+2x-3}\)
\(\Leftrightarrow x^2-x+2x+6=12\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
kl: x thuộc {-3;2}
5.
\(\Leftrightarrow x^2+7-\left(x+4\right)\sqrt{x^2+7}+4x=0\)
Đặt \(\sqrt{x^2+7}=t>0\)
\(\Rightarrow t^2-\left(x+4\right)t+4x=0\)
\(\Delta=\left(x+4\right)^2-16x=\left(x-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{x+4+x-4}{2}=x\\t=\frac{x+4-x+4}{2}=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2+7}=x\left(x\ge0\right)\\\sqrt{x^2+7}=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7=x^2\left(vn\right)\\x^2+7=16\end{matrix}\right.\)
Câu 6 bạn coi lại đề
4.
ĐKXĐ: ...
Đặt \(\sqrt{x+3}=a\ge0\)
\(\Rightarrow x+a=\sqrt{5x^2-a^2}\)
\(\Rightarrow x^2+2ax+a^2=5x^2-a^2\)
\(\Rightarrow2x^2-ax-a^2=0\)
\(\Rightarrow\left(x-a\right)\left(2x+a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=x\\a=-2x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=x\left(x\ge0\right)\\\sqrt{x+3}=-2x\left(x\le0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=x^2\left(x\ge0\right)\\x+3=4x^2\left(x\le0\right)\end{matrix}\right.\)
ĐKXĐ: x>=0; y>=1 ; z>=2.
câu 1:Từ giả thiết ta có:
\(2\sqrt{x}+2\sqrt{y-1}+2\sqrt{z-2}=x+y+z\)
\(\Leftrightarrow x-2\sqrt{x}+1+\left(y-1\right)-2\sqrt{y-1}+1+\left(z-2\right)-2\sqrt{z-2}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
\(\Leftrightarrow\sqrt{x}=1;\sqrt{y-1}=1;\sqrt{z-2}=1\)
Vậy x=1;y=2;z=3.
Có gì ko hiểu bạn cứ bình luận phía dưới :)
a)\(pt\Leftrightarrow\sqrt{3x^2-6x+4}+\sqrt{2x^2-4x+6}+x^2-2x-2=0\)
\(\Leftrightarrow\sqrt{3x^2-6x+4}-1+\sqrt{2x^2-4x+6}-2+x^2-2x+1=0\)
\(\Leftrightarrow\dfrac{3x^2-6x+4-1}{\sqrt{3x^2-6x+4}+1}+\dfrac{2x^2-4x+6-4}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)
\(\Leftrightarrow\dfrac{3\left(x-1\right)^2}{\sqrt{3x^2-6x+4}+1}+\dfrac{2\left(x-1\right)^2}{\sqrt{2x^2-4x+6}+2}+\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1\right)=0\)
Dễ thấy: \(\dfrac{3}{\sqrt{3x^2-6x+4}+1}+\dfrac{2}{\sqrt{2x^2-4x+6}-2}+1>0\)
\(\Rightarrow\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
b)\(\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}=3-4x-2x^2\)
\(pt\Leftrightarrow\sqrt{3x^2+6x+12}+\sqrt{5x^4-10x^2+9}+2x^2+4x-3=0\)
\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x-8=0\)
\(\Leftrightarrow\sqrt{3x^2+6x+12}-3+\sqrt{5x^4-10x^2+9}-2+2x^2+4x+2=0\)
\(\Leftrightarrow\dfrac{3x^2+6x+12-9}{\sqrt{3x^2+6x+12}+3}+\dfrac{5x^4-10x^2+9-4}{\sqrt{5x^4-10x^2+9}+2}+2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x+1\right)^2}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x+1\right)^2\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2\right)=0\)
Dễ thấy: \(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2>0\)
\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
c, ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}-1=2\\\sqrt{2x-1}-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{2x-1}=-1\left(vn\right)\end{matrix}\right.\)
\(\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tm\right)\)
a, ĐKXĐ: \(x\in R\)
\(\sqrt{3x^2}=x+2\)
\(\Leftrightarrow\sqrt{3}\left|x\right|=x+2\)
TH1: \(\sqrt{3}x=x+2\)
\(\Leftrightarrow\left(\sqrt{3}-1\right)x=2\)
\(\Leftrightarrow x=\sqrt{3}+1\)
TH2: \(\sqrt{3}x=-x-2\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)x=-2\)
\(\Leftrightarrow x=1-\sqrt{3}\)
c: =>3x^2+3y^2=39 và 3x^2-2y^2=-6
=>5y^2=45 và x^2=13-y^2
=>y^2=9 và x^2=4
=>\(\left\{{}\begin{matrix}x\in\left\{2;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)
d: \(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{x}=5\\\sqrt{x}-\sqrt{y}=-\dfrac{11}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\\sqrt{y}=1+\dfrac{11}{2}=\dfrac{13}{2}\end{matrix}\right.\)
=>x=1 và y=169/4
b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4-3=1\\-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9-2=7\end{matrix}\right.\)
=>x+1=11/9 và y+4=-11/19
=>x=2/9 và y=-87/19