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\(1,ĐKx\ge5\)
\(\sqrt{\left(x-5\right)\left(x+5\right)}+2\sqrt{x-5}=3\sqrt{x+5}+6\)
\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)
\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)
\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)
2a,ĐK \(x\ge0;x\ne9\)
,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)
\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(M=\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
a) Ta có: \(\left\{{}\begin{matrix}\sqrt{2}x-y=3\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{2}x-y=3\\\sqrt{2}x+2y=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-3y=1\\x+\sqrt{2}y=\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\sqrt{2}-\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{4\sqrt{2}}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{2}}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)
b) Ta có: \(\left\{{}\begin{matrix}\dfrac{x}{2}-2y=\dfrac{3}{4}\\2x+\dfrac{y}{3}=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-8y=3\\2x+\dfrac{1}{3}y=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{25}{3}y=\dfrac{10}{3}\\2x-8y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{2}{5}\\2x=3+8y=3+8\cdot\dfrac{-2}{5}=-\dfrac{1}{5}\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{1}{10}\\y=-\dfrac{2}{5}\end{matrix}\right.\)
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{2x-3y}{4}-\dfrac{x+y-1}{5}=2x-y-1\\\dfrac{x+y-1}{3}+\dfrac{4x-y-2}{4}=\dfrac{2x-y-3}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5\left(2x-3y\right)}{20}-\dfrac{4\left(x+y-1\right)}{20}=\dfrac{20\left(2x-y-1\right)}{20}\\\dfrac{4\left(x+y-1\right)}{12}+\dfrac{3\left(4x-y-2\right)}{12}=\dfrac{2\left(2x-y-3\right)}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y-4x-4y+4=40x-20y-20\\4x+4y-4+12x-3y-6=4x-2y-6\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x-19y+4-40x+20y+20=0\\16x+y-10-4x+2y+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-34x+y=-24\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-102x+3y=-72\\12x+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-114x=-76\\12x+3y=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\12\cdot\dfrac{2}{3}+3y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\3y=4-8=-4\end{matrix}\right.\)
hay \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{4}{3}\end{matrix}\right.\)
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
Bài 2:
a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)
\(=5m^2-2m+9>0\forall m\)
Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m
Bài 1:
ĐKXĐ \(2x\ne y\)
Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)
HPT trở thành
\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)
Bài 1:
\(\sqrt{\left(4-\sqrt{5}\right)^2}+\sqrt{5+2\sqrt{5}+1}\)
\(=\left|4-\sqrt{5}\right|+\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=4-\sqrt{5}+\sqrt{5}+1=5\)
Bài 2:
a: ĐKXĐ: x>=3
\(\sqrt{x-3}=6\)
=>x-3=36
=>x=36+3=39(nhận)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x-3\right)^2}=12\)
=>\(\left|x-3\right|=12\)
=>\(\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài 3:
a: \(P=\left(\dfrac{3-x\sqrt{x}}{3-\sqrt{x}}+\sqrt{x}\right)\cdot\left(\dfrac{3-\sqrt{x}}{3-x}\right)\)
\(=\dfrac{3-x\sqrt{x}+\sqrt{x}\left(3-\sqrt{x}\right)}{3-\sqrt{x}}\cdot\dfrac{3-\sqrt{x}}{3-x}\)
\(=\dfrac{3-x\sqrt{x}+3\sqrt{x}-x}{3-x}\)
\(=\dfrac{-\sqrt{x}\left(x-3\right)-\left(x-3\right)}{-\left(x-3\right)}=\dfrac{\left(x-3\right)\left(\sqrt{x}+1\right)}{x-3}=\sqrt{x}+1\)
b: \(P=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{x+\sqrt{x}}\right):\dfrac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)
\(=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\dfrac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
c: \(A=\sqrt{3x-1}+3\cdot\sqrt{12x-4}-\sqrt{6^2\left(3x-1\right)}+\sqrt{5}\)
\(=\sqrt{3x-1}+6\sqrt{3x-1}-6\sqrt{3x-1}+\sqrt{5}\)
\(=\sqrt{3x-1}+\sqrt{5}\)
d: \(A=\left(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{a+2}{a-2}\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}\cdot\dfrac{a-2}{a+2}\)
\(=\dfrac{2\left(a-2\right)}{a+2}\)
b) Xét phương trình 2 có
(1-x2 )/(1+xy)2 - (x+y)2 - y2 =1
=>(1-x2)/1+2xy+x2y2-x2-2xy-y2 -y2=1
=>(1-x2) /(1-x2 )-y2(1-x2) -y2 =1
=>(1-x2)/(1-x2)(1-y2) -y2=1
=>1/(1-y2) -y2=1
=>1=(1-y2)2
=>1=1-2y2+y4
=>y4-2y2=0
=>y2(y2-2)=0
=>y=0
y2-2=0
=> y=+√2
=> y=-√2
Thay y vào phương trình 1 là ra x
à nhầm ... sửa lại dòng 6
=> 1/(1-y2) - y2=1
=> 1/(1-y2)=1+y2
=> 1=1-y4
=> y=0
=>x=3
=> x=-3
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
\(a,\sqrt{\left(x-1\right)^2-\left(x^2-3\right)}=3\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x^2-3\right)=9\)
\(\Leftrightarrow x^2-2x+1-x^2+3=9\)
\(\Leftrightarrow4-2x=9\)
\(\Leftrightarrow x=\dfrac{-5}{2}\)
\(b,\dfrac{x+3}{x}+\dfrac{x-3}{x-2}=2\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(2x-2\right)}{x\left(x-2\right)}=2\)
\(\Leftrightarrow\left(x-3\right)\left(2x-2\right)=2x\left(x-2\right)\)
\(\Leftrightarrow2x^2-8x+6=2x^2-4x\)
\(\Leftrightarrow-4x=-6\)
\(\Leftrightarrow x=1,5\)
a) \(\sqrt{(x-1)^{2}- x^2-3)}=3\)
\(\Leftrightarrow \sqrt{x^2-2x+1-x^2+3}=3\)
\(\Leftrightarrow \sqrt{4-2x}=3\)
\(\Leftrightarrow 4-2x = 9\)
\(\Leftrightarrow 2x=-5\)
\(\Leftrightarrow x=-2.5\)
Vậy S = {-2.5}