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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
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: \(\dfrac{3}{\sqrt{3x^2+6x+12}+3}+\dfrac{5\left(x-1\right)^2}{\sqrt{5x^4-10x^2+9}+2}+2>0\)
\(\Rightarrow\left(x+1\right)^2=0\Rightarrow x+1=0\Rightarrow x=-1\)
a) Ta có:
\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n}-\sqrt{n+1}}{n-n-1}=-\sqrt{n}+\sqrt{n+1}\)
\(\Rightarrow A=...=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{48}+\sqrt{49}=-1+7=6\)
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
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: \(\sqrt{3}x=-x-2\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)x=-2\)
\(\Leftrightarrow x=1-\sqrt{3}\)
7.
ĐKXĐ: ...
\(\Leftrightarrow10\sqrt{\left(x+1\right)\left(x^2-x+1\right)}=3\left(x^2+2\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow10ab=3\left(a^2+b^2\right)\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3b-a\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-x+1}=3\sqrt{x+1}\\3\sqrt{x^2-x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=9x+9\\9x^2-9x+9=x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-10x-8=0\\9x^2-10x+10=0\end{matrix}\right.\) (casio)
6.
ĐKXĐ: ...
\(\Leftrightarrow2x^2+4=3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow2a^2+2b^2=3ab\)
\(\Leftrightarrow2a^2-3ab+2b^2=0\)
Phương trình vô nghiệm (vế phải là \(5\sqrt{x^3+1}\) sẽ hợp lý hơn)
2
\(M=2y-3x\sqrt{y}+x^2=y-2x\sqrt{y}+x^2+y-x\sqrt{y}\\ =\left(\sqrt{y}-x\right)^2+\sqrt{y}\left(\sqrt{y}-x\right)\\ =\left(\sqrt{y}-x\right)\left(\sqrt{y}-x+\sqrt{y}\right)\\ =\left(\sqrt{y}-x\right)\left(2\sqrt{y}-x\right)\)
b
\(y=\dfrac{18}{4+\sqrt{7}}=\dfrac{18\left(4-\sqrt{7}\right)}{16-7}=\dfrac{72-18\sqrt{7}}{9}=\dfrac{72}{9}-\dfrac{18\sqrt{7}}{9}=8-2\sqrt{7}\\ =7-2\sqrt{7}.1+1=\left(\sqrt{7}-1\right)^2\)
Thế x = 2 và y = \(\left(\sqrt{7}-1\right)^2\) vào M được:
\(M=2\left(\sqrt{7}-1\right)^2-3.2.\sqrt{\left(\sqrt{7}-1\right)^2}+2^2\\ =2\left(8-2\sqrt{7}\right)-6.\left(\sqrt{7}-1\right)+4\\ =16-4\sqrt{7}-6\sqrt{7}+6+4\\ =26-10\sqrt{7}\)
1:
a: =>2x-2căn x+3căn x-3-5=2x-4
=>căn x-8=-4
=>căn x=4
=>x=16
b: \(\Leftrightarrow\left(\sqrt{x}-2\right)\left(x+2\sqrt{x}+4\right)-3\sqrt{x}\left(\sqrt{x}-2\right)=0\)
=>(căn x-2)(x-căn x+4)=0
=>căn x-2=0
=>x=4