Bài 1 : Thực hiện phép tính
a , \(\left(-6,54.\frac{2}{5}\right).2,5\)
b, \(\sqrt{0,01-\sqrt{16}}\)
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a) \(\sqrt{0,01}+\sqrt{0,16}-12^0=0,1+0,4-1=-0,5\)
b) \(\left(\frac{1}{16}\right)^7:\left(\frac{1}{8}\right)^9=\left[\left(\frac{1}{2}\right)^4\right]^7:\left[\left(\frac{1}{2}\right)^3\right]^9=\left(\frac{1}{2}\right)^{28}:\left(\frac{1}{2}\right)^{27}=\frac{1}{2}\)
a)
\(=\sqrt{0,1^2}+\sqrt{0,4^2}-1\)
\(=|0,1|+|0,4|-1\)
\(=0,1+0,4-1=-0,5\)
b)
= \(\left(\left(\frac{1}{2}\right)^4\right)^7:\left(\left(\frac{1}{2}\right)^3\right)^9\)
=\(\left(\frac{1}{2}\right)^{4\cdot7}:\left(\frac{1}{2}\right)^{3\cdot9}\)
= \(\left(\frac{1}{2}\right)^{28}:\left(\frac{1}{2}\right)^{27}\)
\(=\left(\frac{1}{2}\right)^{28-27}\)
\(=\left(\frac{1}{2}\right)^1=\frac{1}{2}\)
a: Ta có: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)
\(=0\)
b: Ta có: \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}\)
\(=5+7-1\)
=11
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
a) \(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
\(=\frac{3}{5}:\frac{-7}{30}+\frac{3}{5}:\frac{-7}{5}\)
\(=\frac{3}{5}\cdot\frac{30}{-7}+\frac{3}{5}\cdot\frac{5}{-7}\)
\(=\frac{3}{5}\left(\frac{-30}{7}+\frac{-5}{7}\right)=\frac{3}{5}\cdot-5=-3\)
b) \(10\cdot\sqrt{0,01}\cdot\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac{1}{6}\sqrt{4}\)
\(=10\cdot\frac{1}{10}\cdot\frac{4}{3}+3\cdot7-\frac{1}{6}\cdot2\)
\(=\frac{4}{3}+21-\frac{2}{6}=22\)
a) \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\frac{\sqrt{3}.\left(\sqrt{3}+2\right)}{\sqrt{3}}+\frac{\sqrt{2}.\left(\sqrt{2}+1\right)}{1+\sqrt{2}}-2+\sqrt{3}\)
\(=\sqrt{3}+2+\sqrt{2}-2+\sqrt{3}\)
\(=2\sqrt{3}+\sqrt{2}\)
b) \(\frac{-3}{2}.\sqrt{9-4\sqrt{5}}+\sqrt{\left(-4\right)^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{5-4\sqrt{5}+4}+\sqrt{4^2.\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{4^2}.\sqrt{\left(1+\sqrt{5}\right)^2}\)
\(=\frac{-3}{2}.\left|\sqrt{5}-2\right|+4.\left|1+\sqrt{5}\right|\)
\(=\frac{-3}{2}.\left(\sqrt{5}-2\right)+4\left(1+\sqrt{5}\right)\)
\(=\frac{-3\sqrt{5}}{2}+3+4+4\sqrt{5}\)
\(=\frac{-3\sqrt{5}}{2}+4\sqrt{5}+7\)
\(=\frac{-3\sqrt{5}}{2}+\frac{8\sqrt{5}}{2}+\frac{14}{2}\)
\(=\frac{-3\sqrt{5}+8\sqrt{5}+14}{2}=\frac{14+5\sqrt{5}}{2}\)
a) \(-0,8\sqrt{\left(-0,125\right)^2}=-0,8.\left|-0,125\right|=-0.8.0,125=-\dfrac{1}{10}\)
b) \(\sqrt{\left(-2\right)^6}=\sqrt{\left(\left(-2\right)^3\right)^2}=\left|\left(-2\right)^3\right|=8\)
c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=2-\sqrt{3}\)
d) \(\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}\)
\(=\left(8+2.4\right)\left(5.25:7\right):\left\{\left[\dfrac{15}{7}+\dfrac{5}{7}\right]:\left[4:\dfrac{8}{9}\right]\right\}\)
\(=10.4\cdot\dfrac{3}{4}:\left\{\dfrac{20}{7}:\dfrac{9}{2}\right\}\)
\(=7.8:\dfrac{40}{63}=12.285\)
\(a,\frac{-5}{9}.\left(\frac{3}{10}-\frac{2}{5}\right)\)
\(=\frac{-5}{9}.\frac{-1}{10}\)
\(=\frac{1}{18}\)
\(b,2^8:2^5+3^3.2-12\)
\(=2^3+9.2-12\)
\(=8+18-12\)
\(=26-12\)
\(=14\)
Câu c,d em chưa học nên không biết làm ạ, mong mọi người thông cảm!!!
Sửa lại câu b
\(=2^3+27.2-12\)
\(=8+54-12\)
\(=62-12\)
\(=50\)