Câu 1: Rút gọn biểu thức:
a.A= (5√5 +2√45 + √5).√5
b.B= 2√2 +1/2√2-4√50
c.C=1/1-√2 + 1/1+√2
Câu 2: Tìm x, biết:
√(8-4x)2 =2
K
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28 tháng 5 2023
\(D=\dfrac{x+1-x+1+4x+2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\)
Khi x=9+4căn 5 thì \(D=\dfrac{4}{8+4\sqrt{5}}=\dfrac{1}{\sqrt{5}+2}=\sqrt{5}-2\)
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: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)
\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)
\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)
\(=-5\left(x^2-16\right)=-5x^2+80\)
AH
Akai Haruma
Giáo viên
29 tháng 12 2023
Lời giải:
a. $P=1+\sqrt{(1-\sqrt{2})^2}=1+|1-\sqrt{2}|=1+\sqrt{2}-1=\sqrt{2}$
b.
$\sqrt{x-1}=3$ (đk: $x\geq 1$)
$\Leftrightarrow x-1=3^2=9$
$\Leftrightarrow x=9+1=10$ (thỏa mãn)
PN
1 tháng 8 2020
\(F=-3\left(x-8\right)\left(2x+1\right)-\left(x+5\right)\left(2-3x\right)-4x\left(x-6\right)\)
\(=-3\left(-3-8\right)\left(-6+1\right)-\left(5-3\right)\left(2+9\right)+12\left(-9\right)\)
\(=-3\left(-11\right)\left(-5\right)-\left(-2\right)11-12.9\)
\(=-165+22-108=22-273=-251\)
\(G=\left(5x-4\right)\left(5-2x\right)-7x\left(x^2-4x+3\right)+\left(x^2-4x\right)\left(7x-2\right)\)
\(=\left(5-4\right)\left(5-2\right)-7\left(1-4+3\right)+\left(1-4\right)\left(7-2\right)\)
\(=3-7.0+5.\left(-3\right)=3-15=-12\)
\(H=\left(-3x+5\right)\left(x-6\right)-\left(x-1\right)\left(x^2-2x+3\right)+\left(x+2\right)\left(x^2-3\right)\)
\(=\left(3+5\right)\left(-1-6\right)-\left(-1-1\right)\left(1+2+3\right)+\left(-1+2\right)\left(1-3\right)\)
\(=8\left(-7\right)-\left(-2\right)6+1\left(-2\right)=-56+12-2=-46\)
PN
1 tháng 8 2020
\(L=5x\left(x-1\right)\left(2x+3\right)-10x\left(x^2-4x+5\right)-\left(x-1\right)\left(x-4\right)\)
\(=-\frac{5}{3}\left(-\frac{4}{3}\right)\left(-\frac{2}{3}+3\right)+\frac{10}{3}\left(\frac{1}{9}+\frac{4}{3}+5\right)-\left(-\frac{4}{3}\right)\left(-\frac{1}{3}-4\right)\)
\(=\frac{20}{9}\left(\frac{7}{3}\right)+\frac{10}{3}\left(\frac{13}{9}+5\right)+\frac{4}{3}\left(-\frac{13}{3}\right)\)
\(=\frac{140}{27}+\frac{10}{3}.\frac{58}{9}-\frac{52}{9}\)
\(=\frac{140}{27}+\frac{580}{27}-\frac{156}{27}=\frac{140+580-156}{27}=\frac{720-156}{27}=\frac{564}{27}\)
\(M=-7x\left(x-5\right)-\left(x-1\right)\left(x^2-x-2\right)+x^2\left(x-3\right)-5x\left(x-8\right)\)
\(=\frac{-7}{2}\left(\frac{1}{2}-5\right)+\frac{\left(\frac{1}{4}-\frac{1}{2}-2\right)}{2}+\frac{1}{4}\left(\frac{1}{2}-3\right)-\frac{5}{2}\left(\frac{1}{2}-8\right)\)
\(=\frac{7}{2}.\frac{9}{2}-\frac{9}{8}-\frac{1}{4}.\frac{5}{2}+\frac{5}{2}.\frac{15}{2}\)
\(=\frac{63}{4}-\frac{9}{8}-\frac{5}{8}+\frac{75}{4}=\frac{138}{4}-\frac{7}{4}=\frac{131}{4}\)
`1a)A=(5sqrt5+2sqrt{45}+sqrt5).sqrt5`
`=(5sqrt{5}+2sqrt{9.5}+sqrt5).sqrt5`
`=(5sqrt5+6sqrt5+sqrt5).sqrt5`
`=12sqrt5*sqrt5=60`
`b)B=2sqrt2+1/(2sqrt2)-4sqrt{50}`
`=2sqrt2+1/(2sqrt2)-4.sqrt{25.2}`
`=2sqrt2+1/(2sqrt2)-20sqrt2`
`=(8+1-80)/(2sqrt2)`
`=(-71)/(2sqrt2)`
`=(-71sqrt2)/4`
`c)=1/(1-sqrt2)+1/(1+sqrt2)`
`=(sqrt2+1)/(1-2)+(sqrt2-1)/(2-1)`
`=-sqrt2-1+sqrt2-1=-2`
`2)sqrt{(8-4x)^2}=2`
`<=>|8-4x|=2`
`<=>|4-2x|=1`
`<=>` \(\left[ \begin{array}{l}4-2x=1\\4-2x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=3\\2x=5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=\dfrac52\end{array} \right.\)
Vậy `S={3/2,5/2}`
\(A=\left(5\sqrt{5}+6\sqrt{5}+\sqrt{5}\right).\sqrt{5}=12\sqrt{5}.\sqrt{5}=12.5=60\)
\(B=2\sqrt{2}+\dfrac{1}{2}\sqrt{2}-10\sqrt{2}=-7,5.\sqrt{2}\)
\(C=\dfrac{1}{1-\sqrt{2}}+\dfrac{1}{1+\sqrt{2}}=\dfrac{1+\sqrt{2}}{-1}+\dfrac{1-\sqrt{2}}{-1}=-1-\sqrt{2}-1+\sqrt{2}=-2\)
Bài 2:
\(\sqrt{\left(8-4x\right)^2}=2\)
*TH1: x < 2
\(\sqrt{\left(8-4x\right)^2}=2\)
\(\Leftrightarrow8-4x=2\Leftrightarrow4x=6\Leftrightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
*TH2: x ≥ 2
\(\sqrt{\left(8-4x\right)^2}=2\)
\(\Leftrightarrow4x-8=2\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)