Bài 1:Tìm x
a)|x|\(\le\)3
b)|x-1|\(\le\)4
Bài 2:Tính
A=4+22+23+24+........+220
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a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =-\dfrac{12}{5}:\dfrac{7}{5}=\dfrac{-12}{7}\)
=>-10<=x<=-12/7
hay \(x\in\left\{-10;-9;-8;-7;-6;-5;-4;-3;-2\right\}\)
b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{1}{8}\)
=>-13/9<=x<=-1/12
hay \(x=-1\)
Bài 3: \(A=\frac{\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Đặt a+b=x;b+c=y;c+a=z
\(A=\frac{\left(x+y\right)\left(y+z\right)\left(z+x\right)}{xyz}\ge\frac{2\sqrt{xy}.2\sqrt{yz}.2\sqrt{zx}}{xyz}=\frac{8xyz}{xyz}=8\)
Dấu = xảy ra khi \(a=b=c=\frac{1}{3}\)
Bài 4: \(A=\frac{9x}{2-x}+\frac{2}{x}=\frac{9x-18}{2-x}+\frac{18}{2-x}+\frac{2}{x}\ge-9+\frac{\left(\sqrt{18}+\sqrt{2}\right)^2}{2-x+x}=-9+\frac{32}{2}=7\)
Dấu = xảy ra khi\(\frac{\sqrt{18}}{2-x}=\frac{\sqrt{2}}{x}\Rightarrow x=\frac{1}{2}\)
a) \(-4\frac{3}{5}\cdot2\frac{4}{23}\le x\le-2\frac{3}{15}:1\frac{6}{15}\)
=> \(-\frac{23}{5}\cdot\frac{50}{23}\le x\le\frac{-33}{15}:\frac{21}{15}\)
=> \(-10\le x\le\frac{-11}{7}\)
=> \(x\in\left\{-10;-9,-8,-7,-6,-5,-4,-3,-2,-1\right\}\)
Bài 1:
a) Ta có \(\left|x\right|\ge0\) (với mọi \(x\))
Mà \(\left|x\right|\le3\)
\(\Rightarrow0\le\left|x\right|\le3\)
\(\Rightarrow\left|x\right|\in\left\{0;1;2;3\right\}\)
\(\Rightarrow x\in\left\{0;1;2;3;-1;-2;-3\right\}\)
b) Ta có: \(\left|x-1\right|\ge0\) (với mọi \(x\))
Mà \(\left|x-1\right|\le4\)
\(\Rightarrow0\le\left|x-1\right|\le4\)
\(\Rightarrow\left|x-1\right|\in\left\{0;1;2;3;4\right\}\)
\(\Rightarrow x-1\in\left\{0;1;-1;2;-2;3;-3;4;-4\right\}\)
\(\Rightarrow x\in\left\{1;2;0;3;-1;4;-2;5;-3\right\}\)
Bài 2:
\(A=4+2^2+2^3+2^4+...+2^{20}\)
\(\Rightarrow A=2+2+2^2+2^3+2^4+...+2^{20}\)
Đặt \(B=2+2^2+2^3+...+2^{20}\)
\(\Rightarrow2B=2^2+2^3+2^4+...+2^{21}\)
\(\Rightarrow2B-B=\left(2^2+2^3+2^4+...+2^{21}\right)-\left(2+2^2+2^3+...+2^{20}\right)\)
\(\Rightarrow B=2^{21}-2\)
\(\Rightarrow A=2+2^{21}-2\)
\(\Rightarrow A=2^{21}\)
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