TK
1
K
Khách
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22 tháng 10 2023
1:
a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)
\(=4x^2-20x+25-4x^2-12x\)
=-32x+25
b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)
\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)
c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)
\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)
\(=\left(-3\right)^2+5\left(2x-3\right)\)
\(=9+10x-15=10x-6\)
2:
a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)
\(=9x^2-12x+4-5x^2+20x+4x-4\)
\(=4x^2+12x\)
b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)
\(=27-x^3+x^3-9x^2+27x-27\)
\(=-9x^2+27x\)
c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
22 tháng 2 2022
b: \(=\dfrac{4x\left(x-1\right)\left(x+1\right)}{6x\left(x-1\right)}=\dfrac{2\left(x+1\right)}{3}\)
c: \(=\dfrac{\left(5-x-1\right)\left(5+x+1\right)}{\left(x+6\right)^2}=\dfrac{\left(4-x\right)\left(x+6\right)}{\left(x+6\right)^2}=\dfrac{4-x}{x+6}\)
d: \(=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
21 tháng 2 2022
a: \(=\dfrac{4x\left(3x+1\right)}{\left(3x+1\right)\left(3x-1\right)}=\dfrac{4x}{3x-1}\)
b: \(=\dfrac{2\left(4x^2-4x+1\right)}{4x-30+2x}=\dfrac{4\left(2x-1\right)^2}{6x-30}=\dfrac{2\left(2x-1\right)^2}{3\left(x-5\right)}\)
d: \(=\dfrac{x\left(x-6\right)}{2\left(x-6\right)\left(x+6\right)}=\dfrac{x}{2x+12}\)
KT
1
22 tháng 2 2021
1) Ta có: \(\dfrac{x\left|x-2\right|}{x^2-5x+6}\)
\(=\left[{}\begin{matrix}\dfrac{-x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x< 2\right)\\\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\left(x>2\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}\dfrac{-x}{x-3}\\\dfrac{x}{x-3}\end{matrix}\right.\)
2) Ta có: \(\dfrac{a^{2x}-b^{2x}}{a^x-b^x}\)
\(=\dfrac{\left(a^x\right)^2-\left(b^x\right)^2}{a^x-b^x}\)
\(=\dfrac{\left(a^x-b^x\right)\left(a^x+b^x\right)}{a^x-b^x}=a^x+b^x\)
H9
HT.Phong (9A5)
CTVHS
26 tháng 10 2023
a) \(\dfrac{2\left(x+1\right)^2}{4x\left(x+1\right)}\left(x\ne0;x\ne-1\right)\)
\(=\dfrac{2\left(x+1\right)^2:2\left(x+1\right)}{4x\left(x+1\right):2\left(x+1\right)}\)
\(=\dfrac{x+1}{2x}\)
b) \(\dfrac{\left(8-x\right)\left(-x-2\right)}{\left(x+2\right)^2}\left(x\ne-2\right)\)
\(=\dfrac{-\left(8-x\right)\left(x+2\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(8-x\right)}{x+2}\)
\(=\dfrac{x-8}{x+2}\)
c) \(\dfrac{2\left(x-y\right)}{y-x}\left(x\ne y\right)\)
\(=\dfrac{2\left(x-y\right)}{-\left(x-y\right)}\)
\(=-2\)
d) \(\dfrac{\left(x+2\right)^2}{2x+4}\left(x\ne-2\right)\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
T
26 tháng 10 2023
ĐKXĐ: \(x\neq0;x\neq-1\)
\(\dfrac{2(x+1)^2}{4x(x+1)}=\dfrac{2(x+1)}{4x}=\dfrac{x+1}{2x}\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(8-x)(-x-2)}{(x+2)^2}=\dfrac{-(8-x)(x+2)}{(x+2)^2}=\dfrac{x-8}{x+2}\)
$---$
ĐKXĐ: \(x\neq y\)
\(\dfrac{2(x-y)}{y-x}=\dfrac{-2(y-x)}{y-x}=-2\)
$---$
ĐKXĐ: \(x\neq-2\)
\(\dfrac{(x+2)^2}{2x+4}=\dfrac{(x+2)^2}{2(x+2)}=\dfrac{x+2}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 3 2023
1.
\(A=\dfrac{2x-9}{\left(x-2\right)\left(x-3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{\left(2x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{2x-9-\left(x^2-9\right)+\left(2x^2-8\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x^2+2x-8}{\left(x-2\right)\left(x-3\right)}=\dfrac{\left(x-2\right)\left(x+4\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\dfrac{x+4}{x-3}\)
b.
\(A=2\Rightarrow\dfrac{x+4}{x-3}=2\Rightarrow x+4=2\left(x-3\right)\)
\(\Rightarrow x=10\) (thỏa mãn)
2.
\(x^4+2x^2y+y^2-9=\left(x^2+y\right)^2-3^2=\left(x^2+y-3\right)\left(x^2+y+3\right)\)
22 tháng 8 2021
a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)
\(=2x^2-x-2x+\dfrac{5}{2}+4x^2\)
\(=6x^2-3x+\dfrac{5}{2}\)
b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)
\(=3x-y-y-x+2x^2-2x\)
\(=2x^2-2y\)
\(a,=\dfrac{\left(x+1\right)^2}{x\left(x+1\right)}=\dfrac{x+1}{x}\\ b,=\dfrac{-\left(x^2-5x-6\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+1\right)\left(x-6\right)}{\left(x+2\right)^2}\)