rút gọn (x - 4).(x - 2) - (x - 1).(x - 3)
cần gấp ai giả hộ mình vs
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HD
9 tháng 12 2016
\(A=\left(\frac{2x+1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}}{\left(x+\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}.\left(3+x\right)}{-2x}-\sqrt{x}\right) \)
\(A=\left(\frac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3+x}{-2\sqrt{x}}-\sqrt{x}\right)\)
\(A=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3+x+2x}{-2\sqrt{x}}\right)\)
\(A=\left(\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}\right).\left(\frac{3x+3}{-2\sqrt{x}}\right)\)
\(A=\frac{1}{\sqrt{x}-1}.\frac{3.\left(x+1\right)}{-2\sqrt{x}}\)
\(A=\frac{3x+3}{-2\sqrt{x}.\left(\sqrt{x}+1\right)}\)
P/s: hình như đề sai hay sao á, thường thì người ta không cho mẫu là 2 số trừ được như ( x - 3x ) đâu
CC
16 tháng 2 2020
a) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)
\(A=\frac{2x-9}{x^2-5x+6}-\frac{x+3}{x-2}-\frac{2x+1}{3-x}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}+\frac{-\left(x+3\right)}{x-2}-\frac{2x+1}{x-3}\)
\(=\frac{2x-9}{\left(x-2\right)\left(x-3\right)}+\frac{-\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\frac{\left(2x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(2x-9\right)-\left(x^2-9\right)+\left(2x^2-3x-2\right)}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{2x-9-x^2+9+2x^2-3x-2}{\left(x-2\right)\left(x-3\right)}=\frac{x^2-x-2}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{x+1}{x-3}\)
b) \(A=\frac{1}{2}\)\(\Leftrightarrow\frac{x+1}{x-3}=\frac{1}{2}\)\(\Leftrightarrow2\left(x+1\right)=x-3\)
\(\Leftrightarrow2x+2=x-3\)\(\Leftrightarrow2x-x=-3-2\)
\(\Leftrightarrow x=-5\)
Vậy \(A=\frac{1}{2}\Leftrightarrow x=-5\)
c) Xem lại đề
NP
1
22 tháng 6 2019
\(A=\left\{2x-3\left(x-1\right)-5\left[x-4\left(3-2x\right)+10\right]\right\}.\left(-2x\right)\)
\(=\left\{2x-3x+3-5\left[x-12+8x+10\right]\right\}.\left(-2x\right)\)
\(=\left\{-x+3-5\left(7x-2\right)\right\}.\left(-2x\right)\)
\(=\left(-x+3-35x+10\right).\left(-2x\right)\)
\(=\left(-36x+13\right).\left(-2x\right)\)
\(=72x^2-26x\)
AH
Akai Haruma
Giáo viên
7 tháng 7 2021
Lời giải:
ĐKXĐ: $x>0; x\neq 1$
\(P=\frac{1}{\sqrt{x}+1}+\frac{x}{\sqrt{x}(1-\sqrt{x})}=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{1-\sqrt{x}+\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}+1)(1-\sqrt{x})}=\frac{x+1}{1-x}\)
b. Khi $x=\frac{1}{\sqrt{2}}$ thì:
\(P=\frac{\frac{1}{\sqrt{2}}+1}{1-\frac{1}{\sqrt{2}}}=3+2\sqrt{2}\)
24 tháng 9 2021
\(P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+7}{x-4}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\left(\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)
\(=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\)
\(=\dfrac{-x+8\sqrt{x}-15+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{-x+8\sqrt{x}-15+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
24 tháng 9 2021
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+1:\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-2\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-5}\\ P=\dfrac{\left(3-\sqrt{x}\right)\left(\sqrt{x}-5\right)+\left(x-4\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{8\sqrt{x}-15-x+x\sqrt{x}-2x-4\sqrt{x}+8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\\ P=\dfrac{x\sqrt{x}-3x+4\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)}\)
NP
4
2 tháng 8 2021
\(\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)\\ =\left(x+2\right)\left(x+2-x+2\right)\\ =4\left(x+2\right)=4x+8\)
23 tháng 9 2021
\(ĐK:x\ge0;x\ne4\\ P=\dfrac{5x+10\sqrt{x}-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5x+10\sqrt{x}-5\sqrt{x}+6+x-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ P=\dfrac{5\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
23 tháng 9 2021
\(P=\dfrac{5\sqrt{x}}{\sqrt{x}-2}-\dfrac{3-\sqrt{x}}{\sqrt{x}+2}+\dfrac{6x}{4-x}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{5\sqrt{x}\left(\sqrt{x}+2\right)-\left(3-\sqrt{x}\right)\left(\sqrt{x}-2\right)-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{5x+10\sqrt{x}+x-5\sqrt{x}+6-6x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{5\sqrt{x}+6}{x-4}\)
11 tháng 11 2018
1, (x2 + 1)(x - 3) - (x - 3)(x2 - 1)
= (x - 3)(x2 + 1 - x2 + 1)
= (x - 3).2
= 2x - 6
2, A = (x - 3)(x + 3) - (x - 2)2
A = x2 - 9 - (x2 - 4x + 4)
A = x2 - 9 - x2 + 4x - 4
A = 4x - 13
EH
11 tháng 11 2018
1 )
\(\left(x^2+1\right)\left(x-3\right)-\left(x-3\right)\left(x^2-1\right).\)
\(=\left(x-3\right)\left(x^2+1-x^2+1\right)\)
\(=\left(x-3\right).2\)
\(=2x-6\)
2
\(\left(x-3\right)\left(x+3\right)-\left(x-2\right)^2\)
=\(=x^2-9-x^2+4x-4\)
\(=4x-13\)
9 tháng 9 2021
Ta có: \(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
9 tháng 9 2021
\(B=\left(\dfrac{2}{\sqrt{x}+2}-\dfrac{\sqrt{x}-5}{x-4}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\\ B=\dfrac{2\sqrt{x}-4-\sqrt{x}+5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\\ B=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}+2}\)
\(\left(x-4\right)\left(x-2\right)-\left(x-1\right)\left(x-3\right)\)
\(=x^2-2x-4x+8-\left(x^2-3x-x+3\right)\)
\(=x^2-2x-4x+8-x^2+3x+x-3\)
\(=-2x+5\)
(x - 4).(x - 2) - (x - 1).(x - 3)
= x2 - 2x - 4x + 8 - ( x2 - 3x - x + 3 )
= x2 - 2x - 4x + 8 - x2 + 3x + x - 3
= 5 - 2x
......