Rút gọn biểu thức
√9/4-√2+√2
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\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)
\(=\sqrt{8-2\cdot\sqrt{8}\cdot2+4}+2\sqrt{2}+1\)
=2căn 2-2+2căn 2+1
=4căn 2-1
\(A=\frac{x^4-5x^2+4}{x^4-10^2+9}=\frac{x^2\left(x^2-5+4\right)}{x^2\left(x^2-10+9\right)}\)
\(=\frac{x^2-1}{x^2-1}=1\)
\(P=\dfrac{4\cdot36^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{4\cdot\left(2^2\cdot3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=-\dfrac{1}{3}\)
\(\dfrac{\sqrt{9-4\sqrt{5}}}{2-\sqrt{5}}=\dfrac{\sqrt{\left(\sqrt{5}-2\right)^2}}{2-\sqrt{5}}=\dfrac{\sqrt{5}-2}{2-\sqrt{5}}=\dfrac{-\left(2-\sqrt{5}\right)}{2-\sqrt{5}}=-1\)
Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
Ta có
A = 5 ( x + 4 ) 2 + 4 ( x – 5 ) 2 – 9 ( 4 + x ) ( x – 4 ) = 5 ( x 2 + 2 . x . 4 + 16 ) + 4 ( x 2 – 2 . x . 5 + 5 2 ) – 9 ( x 2 – 4 2 ) = 5 ( x 2 + 8 x + 16 ) + 4 ( x 2 – 10 x + 25 ) – 9 ( x 2 – 4 2 ) = 5 x 2 + 40 x + 80 + 4 x 2 – 40 x + 100 – 9 x 2 + 144 = ( 5 x 2 + 4 x 2 – 9 x 2 ) + ( 40 x – 40 x ) + ( 80 + 100 + 144 )
= 324
Đáp án cần chọn là: C
\(A=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\\ A=9x^3-16x-9x^3-72+16x\\ A=-72\)
\(A=x\left(3x-4\right)\left(3x+4\right)-9\left(x+2\right)\left(x^2-2x+4\right)+16x\)
\(=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\)
\(=9x^3-16x-9x^3-72+16x=-72\)
Lời giải:
$A=x[(3x)^2-4^2]-9(x^3+2^3)+16x$
$=x(9x^2-16)-9(x^3+8)+16x$
$=9x^3-16x-9x^3-72+16x$
$=-72$
\(A=x\left(3x-4\right)\left(3x+4\right)-9\left(x+2\right)\left(x^2-2x+4\right)+16x\)
\(=9x^3-16x-9x^3-72+16x\)
=-72
\(\sqrt{\dfrac{9}{4}}-\sqrt{2}+\sqrt{2}\\ =\dfrac{3}{2}-\left(\sqrt{2}-\sqrt{2}\right)\\ =\dfrac{3}{2}-0\\ =\dfrac{3}{2}\)