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\(\left(2-3x\right)^3.3^{12}=18^6\)
=>\(\left(2-3x\right)^3.3^{12}=2^6.\left(3^2\right)^6\)
=>\(\left(2-3x\right)^3.3^{12}=2^6.3^{12}\)
=>\(\left(2-3x\right)^3=2^6\)
=>\(\left(2-3x\right)^3=4^3\)
=>2-3x=4
=>3x=-2
=>x=\(\frac{-2}{3}\)
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THEO ĐỀ BÀI TA CÓ
1^2+2^2+3^2+...+10^2=385
MÀ 2^2+4^2+....+20^2=2(1^2+2^2+....+10^2)=2.385=770
VẬY 2^2+2^4+....+20^2=770
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\(\frac{2x}{3}-\frac{1}{6}.x-\frac{x+4}{2}=0\)
\(\Rightarrow\frac{4x}{6}-\frac{x}{6}-\frac{3x+12}{6}=0\)
\(\Rightarrow\frac{4x-x-3x-12}{6}=0\)
\(\Rightarrow0-12=0\Rightarrow-12=0\left(\text{vô lí}\right)\)
Vậy không có x thỏa mãn
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Ta có \(f\left(x\right)=ax^2+bx^4+x+3+11\)
\(=>f\left(-2\right)=a\left(-2\right)^2+b\left(-2\right)^4+\left(-2\right)+3+11=3\)
\(=>f\left(-2\right)=4a+16b+12=3\)
Ta có \(f\left(2\right)=a\left(2^2\right)+b\left(2^4\right)+2+3+11\)
\(=>f\left(2\right)=4a+16b+16\)
\(=>f\left(2\right)=4a+16b+12+4\)
Mà \(f\left(-2\right)=4a+16b+12=3\)
\(=>f\left(2\right)=4a+16b+12+4\)
\(=>f\left(2\right)=3+4\)
\(=>f\left(2\right)=7\)
8,75
5/10