(m^3−m+1)^2+(m^2−3)^2−2(m^2−3)(m^3−m+1) rút gọn biểu thức
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b: A=x+|x-2|
TH1: x>=2
A=x+x-2=2x-2
TH2: x<2
A=x+2-x=2
c: B=|x-3|-x
TH1: x>=3
B=x-3-x=-3
TH2: x<3
B=3-x-x=3-2x
\(=\dfrac{\left(\sqrt{m}+2\right)\left(\sqrt{m}-2\right)}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}-\dfrac{5}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}-\dfrac{\sqrt{m}+3}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}\)
\(=\dfrac{m-4-5-\sqrt{m}-3}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}=\dfrac{m-\sqrt{m}-12}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}\)
\(=\dfrac{\left(\sqrt{m}-4\right)\left(\sqrt{m}+3\right)}{\left(\sqrt{m}+3\right)\left(\sqrt{m}-2\right)}=\dfrac{\sqrt{m}-4}{\sqrt{m}-2}\)
a: \(M=\dfrac{-y+4}{y-2}+\dfrac{1}{y-2}+\dfrac{3}{y+2}\)
\(=\dfrac{-y+5}{y-2}+\dfrac{3}{y+2}=\dfrac{-y^2-2y+5y+10+3y-6}{\left(y-2\right)\left(y+2\right)}\)
\(=\dfrac{-y^2+6y+4}{\left(y-2\right)\left(y+2\right)}\)
b: Khi y=3 thì \(M=\dfrac{-3^2+6\cdot3+4}{\left(3-2\right)\left(3+2\right)}=\dfrac{-5+18}{5}=\dfrac{13}{5}\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
\(3,\\ a,=a^2+2a+1-a^2+2a-1-3a^2+3=-3a^2+4a+3\\ b,=\left(m^3-m+1-m^2+3\right)^2=\left(m^3-m^2-m+4\right)^2\\ 4,\\ a,\Leftrightarrow25x^2+10x+1-25x^2+9=3\\ \Leftrightarrow10x=-7\Leftrightarrow x=-\dfrac{7}{10}\\ b,\Leftrightarrow-9x^2+30x-25+9x^2+18x+9=30\\ \Leftrightarrow48x=46\Leftrightarrow x=\dfrac{23}{24}\\ c,\Leftrightarrow x^2+8x+16-x^2+1=16\\ \Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\)
a, Ta có : \(M=4x^2-9-2\left(x^2+10x+25\right)-2\left(x^2-x+2x-2\right)\)
\(=4x^2-9-2x^2-20x-50-2x^2+2x-4x+4\)
\(=-22x-55\)
b, - Thay \(x=-2\dfrac{1}{3}=-\dfrac{7}{3}\) vào M ta được :
\(M=-\dfrac{11}{3}\)
c, - Thay M = 0 ta được : -22x - 55 = 0
=> x = -2,5
Vậy ...
a) Ta có: \(M=\left(2x+3\right)\left(2x-3\right)-2\left(x+5\right)^2-2\left(x-1\right)\left(x+2\right)\)
\(=4x^2-9-2\left(x^2+10x+25\right)-2\left(x^2+2x-x-2\right)\)
\(=4x^2-9-2x^2-20x-50-2\left(x^2+x-2\right)\)
\(=2x^2-20x-59-2x^2-2x+4\)
\(=-22x-55\)
b) Thay \(x=-2\dfrac{1}{3}\) vào biểu thức \(M=-22x-55\), ta được:
\(M=-22\cdot\left(-2+\dfrac{1}{3}\right)-55\)
\(=-22\cdot\left(\dfrac{-6}{3}+\dfrac{1}{3}\right)-55\)
\(=-22\cdot\dfrac{-5}{3}-55\)
\(=\dfrac{110}{3}-55=\dfrac{110}{3}-\dfrac{165}{3}\)
hay \(M=-\dfrac{55}{3}\)
Vậy: Khi \(x=-2\dfrac{1}{3}\) thì \(M=-\dfrac{55}{3}\)
c) Để M=0 thì -22x-55=0
\(\Leftrightarrow-22x=55\)
hay \(x=-\dfrac{5}{2}\)
Vậy: Khi M=0 thì \(x=-\dfrac{5}{2}\)
\(\left(m^3-m+1\right)^2+\left(m^2-3\right)-2\left(m^2-3\right)\left(m^3-m+1\right)\)
\(=\left(m^3-m+1+m^2-3\right)^2\)
\(=\left(m^3+m^2-m-2\right)^2\)