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23 tháng 5 2022
a: \(=\left(\dfrac{11}{17}+\dfrac{6}{17}\right)+\left(-\dfrac{5}{13}-\dfrac{8}{13}\right)+\dfrac{11}{25}\)
=11/25+1-1=11/25
b: \(=\sqrt{36\cdot\dfrac{1}{4}}+11=9+11=20\)
c: \(=\left(0.25\right)^8\cdot4^8=\left(0.25\cdot4\right)^8=1\)
d: \(=2.8\left(-6.5-3.5\right)=-10\cdot2.8=-28\)
S
24 tháng 12 2022
\(\left(0.25\right)^{10}.4^{10}+\sqrt{5^2-3^2}\)
\(=0.4^{10}+\sqrt{25-9}\)
\(=0+\sqrt{16}=0+4=4\)
\(\dfrac{5}{20}+\dfrac{18}{11}-25\%-\left(\dfrac{18}{11}-\dfrac{4}{9}\right)\)
\(=\dfrac{5}{20}+\dfrac{18}{11}-\dfrac{1}{4}-\dfrac{18}{11}+\dfrac{4}{9}\)
\(=\left(\dfrac{5}{20}-\dfrac{1}{4}\right)+\left(\dfrac{18}{11}-\dfrac{18}{11}\right)+\dfrac{4}{9}\)
\(=0+0+\dfrac{4}{9}=\dfrac{4}{9}\)
BT
23 tháng 2 2018
Ta có: \(A=\left(0,25\right)^{-1}.\dfrac{1}{4}^{-2}.\dfrac{4}{3}^{-2}.\dfrac{5}{4}^{-1}.\dfrac{2}{3}^{-3}\)
--> A= \(\left(\dfrac{\dfrac{1}{1}}{4}\right).\left(\dfrac{\dfrac{1}{1}}{4^2}\right).\left(\dfrac{\dfrac{1}{4^2}}{3^2}\right).\left(\dfrac{\dfrac{1}{5}}{4}\right).\left(\dfrac{\dfrac{1}{2^3}}{3^3}\right)\)
--> A= 4.42. \(\dfrac{3^2}{4^2}\).\(\dfrac{4}{5}\) . \(\dfrac{3^3}{2^3}\)
--> A= \(\dfrac{4.4^2.3^2.3^3}{4^{2.}.5.2^3}=\dfrac{2.3^5}{5}=\dfrac{2.243}{5}\)
--> A= 97,2
23 tháng 5 2022
a: \(=\dfrac{11}{21}+\dfrac{10}{21}+\dfrac{-2}{7}+\dfrac{-5}{7}-\dfrac{6}{5}=\dfrac{-6}{5}\)
b: \(=\left[0.25\cdot\left(-4\right)\right]^5\cdot\left(-\dfrac{50}{9}\right)\)
=50/9
c: \(=\dfrac{3}{22}\left(19+\dfrac{1}{7}+2+\dfrac{6}{7}\right)-6\)
\(=\dfrac{3}{22}\cdot22-6=3-6=-3\)
H
24 tháng 12 2023
\(a,\cdot\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}\right)^2\right]\right\}:\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}\\ =\left[\left(8:2,4\right)\cdot\left(5,25:7\right)\right]:\left[\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)\right]\\ =\left(\dfrac{10}{3}\cdot\dfrac{3}{4}\right):\left(3:\dfrac{9}{2}\right)\\ =\dfrac{5}{2}:\dfrac{2}{3}\\ =\dfrac{15}{4}\)
24 tháng 12 2023
a: \(\dfrac{\left\{\left[\left(2\sqrt{2}\right)^2:2,4\right]\cdot\left[5,25:\left(\sqrt{7}^2\right)\right]\right\}}{\left\{\left[2\dfrac{1}{7}:\dfrac{\left(\sqrt{5}\right)^2}{7}\right]:\left[2^2:\dfrac{\left(2\sqrt{2}\right)^2}{\sqrt{81}}\right]\right\}}\)
\(=\dfrac{\dfrac{8}{2,4}\cdot\dfrac{5,25}{7}}{\left(\dfrac{15}{7}:\dfrac{5}{7}\right):\left(4:\dfrac{8}{9}\right)}\)
\(=\dfrac{\dfrac{10}{3}\cdot\dfrac{3}{4}}{3:\left(4\cdot\dfrac{9}{8}\right)}\)
\(=\dfrac{\dfrac{10}{4}}{3:\left(\dfrac{9}{2}\right)}=\dfrac{5}{2}:\left(3\cdot\dfrac{2}{9}\right)=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{15}{4}\)
b: \(\sqrt{\left(x-\sqrt{2}\right)^2}=\left|x-\sqrt{2}\right|>=0\forall x\)
\(\sqrt{\left(y+\sqrt{2}\right)^2}=\left|y+\sqrt{2}\right|>=0\forall y\)
\(\left|x+y+z\right|>=0\forall x,y,z\)
Do đó: \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|>=0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\sqrt{2}=0\\y+\sqrt{2}=0\\x+y+z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=\sqrt{2}\\y=-\sqrt{2}\\z=0\end{matrix}\right.\)
16 tháng 6 2022
3: |2x-1|=|x+1|
=>2x-1=x+1 hoặc 2x-1=-x-1
=>x=2 hoặc 3x=0
=>x=2 hoặc x=0
4: \(\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{5}=0\\y-\sqrt{3}=0\\x-y-z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\sqrt{5}\\y=\sqrt{3}\\z=x-y=-\sqrt{5}-\sqrt{3}\end{matrix}\right.\)
\((\sqrt{x} -1)^2=(\frac{1}{2})^2\) (ĐK: \(x\geq0\))
TH1: \(\sqrt{x}-1=\frac{1}{2}\)
\(=>\sqrt{x}=\frac{3}{2}\)
\(=> x=\frac{9}{4}\)(t/m)
TH2:\(\sqrt{x}-1=\frac{-1}{2}\)
\(=>\sqrt{x}=\frac{1}{2}\)
\(=> x=\frac{1}{4}\)(t/m)