Bài 1:Cho biểu thức A=|2y-1|+ |x+4y| +5 .Tính giá trị biểu thức A khi:
a,|x|=1/2 và y =1/4
b,|x|=1;|y|=3
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a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)
b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)
Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)
c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)
Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
\(x^2+4y^2-5x+10y-4xy+20\)
\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)
\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)
\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được :
\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)
\(B=x^2-2xy-2x+2y+y^2\)
\(=x^2-2xy+y^2-2\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được :
\(=1-2=-1\)
a)\(A=\left(\frac{x+y}{x-2y}+\frac{3y}{2y-x}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x+y-3y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(\frac{x-2y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)
\(=\left(1-3xy\right).\frac{-x-1}{1-3xy}+\frac{x^2}{x+1}\)
\(=-\left(x+1\right)+\frac{x^2}{x+1}\)`
\(=\frac{-\left(x+1\right)^2+x^2}{x+1}\)
\(=\frac{-x^2-2x-1+x^2}{x+1}\)
\(=\frac{-2x-1}{x+1}\)(1)
b) Thay \(x=-3,y=2014\)vào (1) ta được:
\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)
Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014
a: \(A=x^2+2xy+y^3=5^2+2\cdot5\cdot4+4^3=129\)
b: \(B=\left(-1\right)\cdot\left(-1\right)-\left(-1\right)^2\cdot\left(-1\right)^2+\left(-1\right)^4\cdot\left(-1\right)^4-\left(-1\right)^6\cdot\left(-1\right)^6=1-1+1-1=0\)
\(1)A=2x\left(x-y\right)-y\left(y-2x\right)\)
\(=2x^2-2xy-y^2+2xy\)
\(=2x^2-y^2=2.\left(-\dfrac{2}{3}\right)^2-\left(-\dfrac{1}{3}\right)^2\)
\(=\dfrac{8}{9}-\dfrac{1}{9}=\dfrac{7}{9}\)
\(2)B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(=5x^2-20xy-4y^2+20xy\)
\(=5x^2-4y^2=5.\left(-\dfrac{1}{5}\right)^2-4.\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(3)C=\text{x.(x^2-y^2)-x^2(x+y)+y(x^2-x)}\)
\(=x^3-xy^2-x^3-x^2y+x^2y-xy\)
\(=-xy\left(x+1\right)\)
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
ai xong trước mk sẽ k cho