rút gọn B=(√x/x√x-1 + 1/√x-1):√x + 1/x+√x +1
AE
Những câu hỏi liên quan
1. Rút gọn biểu thức A = \(\dfrac{\text{√ x + 1}}{\text{√ x − 1 }}-\dfrac{\text{√ x − 1}}{\text{√ x + 1}}+\dfrac{\text{8 √ x}}{\text{1 − x }}\)
2. Rút gọn biểu thức B = \(\dfrac{\text{√ x − x − 3}}{\text{x − 1 }}-\dfrac{\text{1}}{\text{√ x − 1 }}\) với x ≥ 0, x ≠ 1
Rút gọn a, (x-1) (x²+x²+x+1)
b, (x+1) (x^4 -x³+x²-x+1)
Xem chi tiết
a,(x-1) (x²+x²+x+1)
=(x-1)(2x2+x+1)
=2x3+2x+x-2x2-x-1
=2x3-2x2+2x-1
b, (x+1) (x4 -x3+x2-x+1)
=x5-x4+x3-x2+x+x4-x3+x2-x+1
=x5+1
2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
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HH
8 tháng 8 2023 lúc 12:12
Rút gọn biểu thức
a) A= {[(x√x)-1]\[(√x)-1]+√x}{[(x√x)+1]/[(√x)+1]-√x}
b) B={[ 3/x-(√x)-2] +1/[(√x)+1]}.[(√x)-2]
a) \(A=\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(A=\left[\dfrac{\left(\sqrt{x}\right)^3-1^3}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}\right)^3+1^3}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}-1}+\sqrt{x}\right]\left[\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\left(x+\sqrt{x}+1+\sqrt{x}\right)\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(A=\left(x+2\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)\)
\(A=\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)^2\)
\(A=\left[\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\right]^2\)
\(A=\left(x-1\right)^2\)
\(A=x^2+2x+1\)
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Rút gọn a)(2x-1)² - (4x+1)(x-2)
b)(x-3)³ - (x+1)(x²-x+1)
\(\left(2x-1\right)^2-\left(4x+1\right)\left(x-2\right)\\ =\left[\left(2x\right)^2-2\cdot2x\cdot1+1^2\right]-\left(4x^2-8x+x-2\right)\\ =4x^2-4x+1-4x^2+8x-x+2\\ =3x+3\)
\(\left(x-3\right)^3-\left(x+1\right)\left(x^2-x+1\right)\\ =\left(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3\right)-\left(x^3-x^2+x+x^2-x+1\right)\\ =x^3-9x^2+27x-27-x^3+x^2-x-x^2+x-1\\ =-10x^2+27x-28\)
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1 a..Rút gọn biểu thức A = \(\dfrac{\text{ x 2 − 4 x + 4}}{\text{x 3 − 2 x 2 − ( 4 x − 8 ) }}\)
b. Rút gọn biểu thức B = \(\left(\dfrac{x+2}{\text{x }\sqrt{\text{x }}+1}-\dfrac{1}{\sqrt{\text{x}}+1}\right).\dfrac{\text{4 }\sqrt{x}}{3}\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
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\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
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Rút gọn biểu thức B: \(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\)
Ta có: \(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{4}{x^2-1}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4x+4}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-4}{x+1}\)
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\(\dfrac{x-1}{x+1}\) - \(\dfrac{x+1}{-\left(1-x\right)}\) - \(\dfrac{4}{\left(1-x\right)\left(1+x\right)}\) MTC: -(1 - x)(1 + x)
= \(\dfrac{-\left(x-1\right)^2}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{\left(x+1\right)^2}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{-4}{\text{-(1 - x)(1 + x)}}\)
= \(\dfrac{-x^2+2x-1}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{x^2+2x+1}{\text{-(1 - x)(1 + x)}}\) - \(\dfrac{-4}{\text{-(1 - x)(1 + x)}}\)
= \(\dfrac{-2x^2+2}{\text{-(1 - x)(1 + x)}}\) = \(\dfrac{-2\left(x^2-1\right)}{\text{-(1 - x)(1 + x)}}\) = \(\dfrac{2\text{(x - 1)(1 + x)}}{\text{(1 - x)(1 + x)}}\) = \(\dfrac{2x-2}{1-x}\)
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rút gọn biểu thức B=x(1-x^2)^2/1+x^2:((1-x^3/1-x+x)(1+x^3/1+x-x))
rút gọn đa thức:
a(x+1)(x-1)
b(x+1)(x-1)(x2+1)
c(x+1)(x-1)(x2+1)(x4+1)-x8
a) \(\left(x+1\right)\left(x-1\right)\)
\(=x^2-1^2\)
\(=x^2-1\)
b) \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\)
\(=\left(x^2\right)^2-1^2\)
\(=x^4-1\)
c) \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)\left(x^2+1\right)-x^8\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)-x^8\)
\(=\left(x^4-1\right)\left(x^4+1\right)-x^8\)
\(=\left(x^4\right)^2-1-x^8\)
\(=x^8-1-x^8\)
\(=-1\)
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