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B1: a) \(\left|x-2\right|+9y^2+12xy+4x^2=0\)
=> \(\left|x-2\right|+\left(3y+2x\right)^2=0\)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\left(3y+2x\right)^2\ge0\forall x;y\)
=> \(\left|x-2\right|+\left(3y+2x\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}x-2=0\\3y+2x=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\3y=-2x\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\3y=-2.2=-4\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2\\y=-\frac{4}{3}\end{cases}}\)
Vậy ...
\(\left(2^2+4^2+6^2+...+100^2\right)-\left(1^2+3^2+5^2+99^2\right)\\ =\left(2^2-1^2\right)+\left(4^2-3^2\right)+\left(6^2-5^2\right)+...+\left(100^2-99^2\right)\\ =\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+\left(6-5\right)\left(6+5\right)+...+\left(100-99\right)\left(100+99\right)\\ =3+7+11+...+199=\frac{50.202}{2}=5050\)
a. 134^2 - 68.134 + 34^2 = ( 134 - 34 ) ^2 = 100^2 = 10000
b. 9^8.2^8 - ( 18^4 - 1 )(18^4 + 1 ) = 18^8 - 18^8 + 1 = 1
c. 100^2 - 99^2 + 98^2 - 97^2 + ... + 2^2 - 1
=( 100 - 99 )( 100 + 99 ) + ( 98 - 97 )( 98 + 97 ) + ... + ( 2 - 1 )( 2 + 1 )
= 100 + 99 + 98 + 97 + ... + 2 + 1
=( 100 + 1 ).100:2 = 5050
a: \(\Leftrightarrow x\left(16-x^2\right)+x^3-125=3\)
=>16x-125=3
=>16x=128
hay x=8
b: \(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
=>12x-4=-10
=>12x=-6
hay x=-1/2
c: \(\Leftrightarrow x^3-27+x\left(4-x^2\right)=1\)
\(\Leftrightarrow4x-27=1\)
hay x=7
Bài 11:
1) Sửa lại đề là: \(A=127^2+146.127+73^2\)
\(\Rightarrow A=127^2+2.127.73+73^2\)
\(\Rightarrow A=\left(127+73\right)^2\)
\(\Rightarrow A=200^2\)
\(\Rightarrow A=40000\)
Vậy \(A=40000.\)
2) Sửa lại đề là: \(B=9^8.2^8-\left(18^4-1\right).\left(18^4+1\right)\)
\(\Rightarrow B=\left(9.2\right)^8-\left[\left(18^4\right)^2-1^2\right]\)
\(\Rightarrow B=18^8-\left(18^8-1\right)\)
\(\Rightarrow B=18^8-18^8+1\)
\(\Rightarrow B=0+1\)
\(\Rightarrow B=1\)
Vậy \(B=1.\)
4) \(D=\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(\Rightarrow2D=\left(3-1\right).\left(3+1\right).\left(3^2+1\right).\left(3^4+1\right).\left(3^8+1\right).\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(\Rightarrow D=\frac{3^{32}-1}{2}\)
1272 + 146.127 + 732
= 1272 + 2 . 73 .127 + 732
= (127 + 73 ) 2
= 200 2
a) 2 +4+6+8+...+2018
= ( 2018+2) x 1009 : 2
= 2020 x 1009 : 2
= 1009 x (2020:2)
= 1009 x 1010
= 1 019 090
b) S = 10 + 102 + 103 + ...+ 10100
=> 10.S = 102 + 103 + 104 +...+ 10101
=> 10.S - S = 10101-10
9.S=10101- 10
\(\Rightarrow S=\frac{10^{101}-10}{9}\)
c) \(S=\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5S=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5S-S=1-\frac{1}{5^{100}}\)
\(4S=1-\frac{1}{5^{100}}\)
\(S=\frac{1-\frac{1}{5^{100}}}{4}\)
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d) \(S=\frac{1!}{3!}+\frac{2!}{4!}+\frac{3!}{5!}+...+\frac{2018!}{2020!}\)
\(S=\frac{1}{1.2.3}+\frac{1.2}{1.2.3.4}+\frac{1.2.3}{1.2.3.4.5}+...+\frac{1.2.3...2018}{1.2.3...2020}\)
\(S=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2019.2020}\)
\(S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(S=\frac{1}{2}-\frac{1}{2020}\)
\(S=\frac{1009}{2020}\)
\(A=-1^2+2^2-3^2+4^2-...-99^2+100^2\)
\(A=\left(2-1\right).\left(1+2\right)+\left(4-3\right).\left(3+4\right)+...\left(+100-99\right).\left(99+100\right)\)
\(A=1.\left(1+2+3+...+99+100\right)\)
\(A=\dfrac{100.\left(100+1\right)}{2}=50.101=5050\)
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