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23 tháng 12 2021
e: \(E=\dfrac{x^2-9-x^2+4-x^2+9}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x+2}{x+3}\)
a: \(A=\dfrac{4x^2+x^2-2x+1+x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{6x^2+2}{\left(x-1\right)\left(x+1\right)}\)
21 tháng 7 2021
Bài 8
a, \(A=a^2+b^2=\left(a+b\right)^2-2ab\Rightarrow S^2-2P\)
b, \(B=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\)
\(\Rightarrow S\left(S^2-3P\right)=S^3-3PS\)
c, \(C=a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2\)
\(\Rightarrow\left(S^2-2P\right)^2-2P^2\)
HP
14 tháng 8 2021
\(\dfrac{2x-3}{5}-x+2\ge\dfrac{x}{3}\)
\(\Leftrightarrow3\left(2x-3\right)-15\left(x+2\right)\ge5x\)
\(\Leftrightarrow6x-9-15x+30\ge5x\)
\(\Leftrightarrow6x-15x-5x\ge9+30\)
\(\Leftrightarrow-14x\ge-21\)
\(\Leftrightarrow x\le\dfrac{21}{14}\le\dfrac{3}{2}\)
-------------|--------]////////////////--->
0 3/2
lâu rồi cũng không nhớ cách làm :v
A
1
24 tháng 2 2022
\(=\dfrac{\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x-y\right)}{\left(x-y\right)^2x\left(x+y\right)}=\dfrac{\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)}{x\left(x-y\right)^2\left(x+y\right)}=\dfrac{x^2+y^2}{x}\)
11 tháng 1 2022
Bài 1:
b: \(\Leftrightarrow x^2-2x+4+x^3+8=12\)
\(\Leftrightarrow x^3+x^2-2x=0\)
=>x(x+1)=0
=>x=0 hoặc x=-1
f: \(\Leftrightarrow x+3-6x+12=-5\)
=>-5x=-20
hay x=4(nhận)
27 tháng 8 2021
a) \(x^2+2x+1=\left(x+1\right)^2\)
\(x^2-2x+1=\left(x-1\right)^2\)
\(x^2+4x+4=\left(x+2\right)^2\)
\(x^2-4x+4=\left(x-2\right)^2\)
\(x^2+6x+9=\left(x+3\right)^2\)
\(x^2-6x+9=\left(x-3\right)^2\)
\(x^2-10x+25=\left(x-5\right)^2\)
\(x^2+10x+25=\left(x+5\right)^2\)
b) \(16x^2-8x+1=\left(4x-1\right)^2\)
c) \(4x^2+12xy+9y^2=\left(2x+3y\right)^2\)
d) \(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
e) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
f) \(9x^2+30x+25=\left(3x+5\right)^2\)
14 tháng 12 2021
\(a,PT\left(1\right)=\dfrac{75y^4}{42x^2y^5};PT\left(2\right)=\dfrac{28x}{42x^2y^5}\\ b,PT\left(1\right)=\dfrac{11y^2}{102x^4y^3};PT\left(2\right)=\dfrac{9x^3}{10x^4y^3}\\ c,PT\left(1\right)=\dfrac{3x\left(3x+1\right)}{36x^2y^4};PT\left(2\right)=\dfrac{4y\left(y-2\right)}{36x^2y^4}\\ d,PT\left(1\right)=\dfrac{6y^2}{36x^3y^4};PT\left(2\right)=\dfrac{4x\left(x+1\right)}{36x^3y^4};PT\left(3\right)=\dfrac{9x^2y\left(x-1\right)}{36x^3y^4}\)
\(e,PT\left(1\right)=\dfrac{12y^4\left(3+2x\right)}{120x^4y^5};PT\left(2\right)=\dfrac{75x^2y^3}{120x^4y^5};PT\left(3\right)=\dfrac{8x^3}{120x^4y^5}\\ f,PT\left(1\right)=\dfrac{3\left(x+1\right)\left(4x-4\right)}{6x\left(x+3\right)\left(x+1\right)};PT\left(2\right)=\dfrac{2\left(x+3\right)\left(x-3\right)}{6x\left(x+1\right)\left(x+3\right)}\)
14 tháng 12 2021
\(g,PT\left(1\right)=\dfrac{4x^2}{2x\left(x+2\right)^3};PT\left(2\right)=\dfrac{\left(x-2\right)\left(x+2\right)}{2x\left(x+2\right)^3}\\ h,PT\left(1\right)=\dfrac{5}{3x\left(x-2\right)\left(x+2\right)}=\dfrac{10\left(x+3\right)}{6x\left(x-2\right)\left(x+2\right)\left(x+3\right)}\\ PT\left(2\right)=\dfrac{3}{2\left(x+2\right)\left(x+3\right)}=\dfrac{9x\left(x-2\right)}{6x\left(x+2\right)\left(x+3\right)\left(x-2\right)}\)
a, \(x>2\)
b, \(x\ne-2;x\ge2\\ \)
c, \(x\ne-2;x>2\)
d, \(3-2x>0\\ x< \dfrac{3}{2}\\ e,2x+3>0\\ x>-\dfrac{3}{2}\\ f,x+1< 0\\ x< -1\)