tính A = 2 mũ 0 + 2 mũ 2 + 2 mũ + 2 mũ 3 + .... + 2 mũ 19 giùm mik đc ko
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1. \(x^4-2x^2+1=\left(x^2-1\right)^2\)
2. \(x^2+5x+\dfrac{25}{4}=x^2+2.x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)
3. \(16x^2-8x+1=\left(4x-1\right)^2\)
4. \(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x-y+1\right)\left(x+y\right)\)
5. \(\dfrac{1}{4}x^2-\dfrac{4}{9}y^2=\left(\dfrac{1}{2}x-\dfrac{2}{3}y\right)\left(\dfrac{1}{2}x+\dfrac{2}{3}y\right)\)
6. \(a^2-2ab+b^2-x^2=\left(a-b\right)^2-x^2=\left(a-b-x\right)\left(a-b+x\right)\)
7. \(4x^2-20x+25-y^2=\left(2x-5\right)^2-y^2=\left(2x-5-y\right)\left(2x-5+y\right)\)
Bài 2:
a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
=>(x+5)(x-6)=0
=>x=-5 hoặc x=6
b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
=>-4x+2=0
hay x=1/2
c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
Ta có A = 1 + 2 + 22 + 23 + ... + 219
=> 2A = 2 + 22 + 23 + 24 + ... + 220
=> 2A - A = (2 + 22 + 23 + 24 + ... + 220) - (1 + 2 + 22 + 23 + ... + 219)
=> A = 220 - 1
Lại có B = 220
=> A và B là 2 số tự nhiên liên tiếp
Ta có: \(A=2^0+2^1+2^2+2^3+...+2^{19}\)
\(\Leftrightarrow2A=2^1+2^2+2^3+2^4...+2^{20}\)
\(\Leftrightarrow2A-A=\left(2^1+2^2+2^3+2^4...+2^{20}\right)-\left(2^0+2^1+2^2+2^3+...+2^{19}\right)\)
\(\Leftrightarrow A=2^{20}-1\)
Vì \(2^{20}-1\)và \(2^{20}\)là 2 STN liên tiếp
\(\Rightarrow\)\(A\)và \(B\)là 2 STN liên tiếp
\(\left(9^{30}-27^{19}\right):3^{57}+\left(125^9-25^{12}\right):5^{24}\)
\(=\left(3^{60}-3^{57}\right):3^{57}+\left(5^{27}-5^{24}\right):5^{24}\)
\(=3^{57}\left(3^3-1\right):3^{57}+5^{24}\left(5^3-1\right):5^{24}\)
\(=3^3-1+5^3-1\)
\(=27-1+125-1\)
\(=150\)
2 )
\(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)-\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=6\end{matrix}\right.\)
Vậy ...
b )
\(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
c )
\(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=-4\left(L\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
a: \(=\left\{145-\left[130-10\right]:2\right\}\cdot5\)
\(=\left\{145-60\right\}\cdot5=85\cdot5=425\)
b: \(=100:\left\{250:\left[450-4\cdot125+4\cdot25\right]\right\}\)
\(=\dfrac{100}{250:\left[450-500+100\right]}=\dfrac{100}{250:50}=\dfrac{100}{5}=20\)
c: \(=355-5\cdot\left[64-\left(27-25\right)\right]=355-5\cdot\left[64-2\right]\)
\(=355-310=45\)
Lời giải:
$A=2^0+2^1+2^2+...+2^{19}$
$2A=2^1+2^2+2^3+...+2^{20}$
$\Rightarrow 2A-A=2^{20}-2^0=2^{20}-1$
$\Rightarrow A=2^{20}-1$