a) \(\left(x+1\right)^3=-27\)
b) \(\sqrt{36}-\sqrt{9}.x=\sqrt{64}\)
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13 tháng 5 2022
\(a,\) ta có :
\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)
\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)
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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :
\(B=\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\left(\sqrt{x}-1\right)+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\sqrt{x}+1+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)
vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)
để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
vậy để \(B=2\) thì \(x=4\)
6 tháng 3 2022
Câu 1 :
a, \(=8+4-2.6=12-12=0\)
b, đk : x > 0 ; x khác 1
\(P=\left(\dfrac{\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right).\dfrac{x+\sqrt{x}}{1-\sqrt{x}}=\dfrac{1-\sqrt{x}}{1-\sqrt{x}}=1\)
AO
16 tháng 12 2017
\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)
\(A=6\sqrt{3^2.3}-2\sqrt{5^2.3}-\frac{1}{2}\sqrt{10^2.3}\)
\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)
\(A=3\sqrt{3}\)
vậy \(A=3\sqrt{3}\)
\(B=\left(1+\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\frac{x-\sqrt{x}}{\sqrt{x}-1}\right)\) \(ĐKXĐ:x>0;x\ne1\)
\(B=\left[1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\)
\(B=\left[1+\sqrt{x}\right]\left[1-\sqrt{x}\right]\)
\(B=1-x\)
vậy \(B=1-x\)
\(C=\sqrt[3]{64}-\sqrt[3]{-125}+\sqrt[3]{216}\)
\(C=\sqrt[3]{4^3}-\sqrt[3]{\left(-5\right)^3}+\sqrt[3]{6^3}\)
\(C=4+5+6\)
\(C=15\)
vậy \(C=15\)
16 tháng 12 2017
Cho mk giải câu a:
\(A=6\sqrt{27}-2\sqrt{75}-\frac{1}{2}\sqrt{300}\)
\(A=18\sqrt{3}-10\sqrt{3}-\frac{1}{2}10\sqrt{3}\)
\(A=18\sqrt{3}-10\sqrt{3}-10:2\sqrt{3}\)
\(A=18\sqrt{3}-10\sqrt{3}-5\sqrt{3}\)
\(A=\left(18-10-5\right)\sqrt{3}\)
\(A=3\sqrt{3}\)
BT
29 tháng 5 2017
\(\sqrt{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=50\)
Vậy x = 50
b) \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
\(\Leftrightarrow\left(x+1\right)\sqrt{3}=2\sqrt{3}+3\sqrt{3}\)
\(\Leftrightarrow\left(x+1\right)\sqrt{3}=\left(2+3\right)\sqrt{3}\)
\(\Leftrightarrow x+1=5\)
\(\Leftrightarrow x=4\)
Vậy x = 4
29 tháng 5 2017
\(\sqrt{9\left(x-1\right)}=21\\9\left(x-1\right)=21^2\\x-1=49\\ x=48 \)\(\sqrt{3}x+\sqrt{3}=2\sqrt{3}+3\sqrt{3}\\ 0=\sqrt{3}\left(2+3-1-x\right)\\ 0=\sqrt{3}\left(4-x\right)\\ x=4\\ \)
NV
Nguyễn Việt Lâm
Giáo viên
29 tháng 12 2021
\(=\left(\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right).\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
30 tháng 8 2021
b: Ta có: \(\sqrt[3]{-0.008}-\dfrac{1}{5}\cdot\sqrt[3]{64}+5\cdot\sqrt[3]{\left(-5\right)^3}\)
\(=-\dfrac{1}{5}-\dfrac{1}{5}\cdot4+5\cdot\left(-5\right)\)
\(=-\dfrac{1}{5}-\dfrac{4}{5}-25\)
=-26
a)\(\left(x+1\right)^3=-27\)
\(\left(x+1\right)^3=\left(-3\right)^3\)
x+1=-3
x=(-3)-1
x=-4
b)6-3x=8
3x=6-8
3x=(-2)
x=\(-\frac{2}{3}\)
a) \(\left(x+1\right)^3=-27\)
\(\Rightarrow\left(x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-4\)
Vậy \(x=-4\)
b) \(\sqrt{36}-\sqrt{9}.x=\sqrt{64}\)
\(\Rightarrow6-3.x=8\)
\(\Rightarrow3x=-2\)
\(\Rightarrow x=\frac{-2}{3}\)
Vậy \(x=\frac{-2}{3}\)