Rút gọn
a,\(|x-2|-|x-3|\)
K
Khách
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Những câu hỏi liên quan
12 tháng 12 2023
a: (x+1)(3-x)(x-2)2
\(=\left(3x-x^2+3-x\right)\left(x^2-4x+4\right)\)
\(=\left(-x^2+2x+3\right)\left(x^2-4x+4\right)\)
\(=-x^4+4x^3-4x^2+2x^3-8x^2+8x+3x^2-12x+12\)
\(=-x^4+6x^3-9x^2-4x+12\)
b: \(9x\left(1-x\right)+\left(3x-2\right)\left(3x+2\right)\)
\(=9x-9x^2+\left(3x\right)^2-4\)
\(=9x-9x^2+9x^2-4=9x-4\)
SS
2
H9
HT.Phong (9A5)
CTVHS
16 tháng 7 2023
a) \(\left(x+3\right)^2+\left(x-3\right)^2+2\left(x^2+9\right)\)
\(=\left(x+3\right)^2+2\left(x+3\right)\left(x-3\right)+\left(x-3\right)^2\)
\(=\left[\left(x+3\right)+\left(x-3\right)\right]^2\)
\(=\left(x+3+x-3\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
b) \(\left(4x-1\right)^3-\left(4x-3\right)\left(16x^2+3\right)\)
\(=\left(64x^3-48x^2+12x-1\right)-\left(64x^3+12x-48x^2-9\right)\)
\(=64x^3-48x^2+12x-1-64x^3-12x+48x^2+9\)
\(=\left(64x^3-64x^3\right)-\left(48x^2-48x^2\right)+\left(12x-12x\right)-\left(1-9\right)\)
\(=0-0+0+8\)
\(=8\)
KV
16 tháng 7 2023
a) (x + 3)² + (x - 3)² + 2(x² - 9)
= (x + 3)² + 2(x + 3)(x - 3) + (x - 3)²
= (x + 3 + x - 3)²
= (2x)²
= 4x²
b) (4x - 1)³ - (4x - 3)(16x² + 3)
= 64x³ - 48x² + 12x - 1 - 64x³ - 12x + 48x² + 9
= (64x³ - 64x³) + (-48x² + 48x²) + (12x - 12x) + (-1 + 9)
= 8
Y
23 tháng 3 2023
\(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}-3}+\dfrac{x}{3\sqrt{x}-x}\right).\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\left(dkxd:x\ne0;\pm\sqrt{3}\right)\)
\(=\left(\dfrac{2}{\sqrt{x}-3}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-3\right)\)
\(=\left(\dfrac{2\sqrt{x}-x}{\sqrt{x}\left(\sqrt{x}-3\right)}\right).\left(\sqrt{x}-3\right)\)
\(=\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}}\)
\(=2-\sqrt{x}\)
Vậy \(A=2-\sqrt{x}\)
11 tháng 7 2021
Bài 1:
a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)
b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)
Bài 2:
a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)
b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)
HT
1
MT
8 tháng 7 2015
nếu chia ra như ông thì A= (x+y+z)^3 - (x+y-z)^3-[(y+z-x)^3 - (z+x-y)^3 ]
=(x+y+z)^3 - (x+y-z)^3-(y+z-x)^3 +(z+x-y)^3 đâu đúng chứ
15 tháng 11 2021
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
29 tháng 6 2021
Ta có: \(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x}+1}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}\)
\(=\dfrac{8+x\left(1+\sqrt{x}-1\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x\sqrt{x}+8}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}+2}{x-4}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{2\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}+4+x-2\sqrt{x}}{2\left(x-4\right)}\)
\(=\dfrac{x+4}{2x-8}\)
LN
1
H9
HT.Phong (9A5)
CTVHS
26 tháng 6 2023
\(A=\dfrac{x\sqrt{x}+1}{x-1}-\dfrac{x-1}{\sqrt{x}+1}\)
\(A=\dfrac{x\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{x\sqrt{x}+1-\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{x\sqrt{x}+1-\left(x\sqrt{x}-x-\sqrt{x}+1\right)}{x-1}\)
\(A=\dfrac{x\sqrt{x}+1-x\sqrt{x}+x+\sqrt{x}-1}{x-1}\)
\(A=\dfrac{x+\sqrt{x}}{x-1}\)
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