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Tìm x hả cậu?
2x+2x+2=160
=>2x(1+22)=160
=>2x.5=160
=>2x=160:5
=>2x=32
=>2x=25
=>x=5
Vậy x=5
\(2^x+2^{x+2}=160\)
\(2^x+2^x\cdot4=160\)
\(2^x\cdot\left(1+4\right)=160\)
\(2^x=160:5=32\)
\(2^x=2^5\)
\(x=5\)
Nhân phân phối là ra thôi
a)
\(VT=\left(x-1\right)\left(x+1\right)=x.x+x.1-1.x+\left(-1\right).1\)
\(=\left(x^2-1\right)+\left(x-x\right)=x^2-1+0=x^2-1=VP\Rightarrow dccm\)
c) thay vì c/m A=B ta chứng Minh B=A
\(VP=\left(x+1\right)\left(x^2-x+1\right)=\left(x^3-x^2+x\right)+\left(x^2-x+1\right)\)
\(=\left(x^3+1\right)+\left(-x^2+x^2\right)+\left(x-x\right)=x^3+1+0+0=x^3+1=VT\Rightarrow VT=VP\Rightarrow dpcm\)\(=x^3+1+0+0=x^3+1=VT\Rightarrow VT=VP\Rightarrow dpcm\)
a: \(A=3\cdot\dfrac{1}{8}\cdot\dfrac{-1}{3}+6\cdot\dfrac{1}{8}\cdot\dfrac{1}{9}+3\cdot\dfrac{1}{2}\cdot\dfrac{-1}{27}\)
\(=\dfrac{-1}{8}+\dfrac{1}{12}-\dfrac{1}{18}=-\dfrac{7}{72}\)
b: \(B=\left(-1\cdot3\right)^2+\left(-1\right)\cdot3-1+27\)
\(=9-3-1+27\)
=36-4=32
c: \(C=-0.7xy^2-2x^2y-4.5xy\)
\(=-0.7\cdot\dfrac{1}{2}\cdot1-2\cdot0.25\cdot\left(-1\right)-4.5\cdot0.5\cdot\left(-1\right)\)
\(=\dfrac{-7}{20}+\dfrac{1}{2}+\dfrac{9}{2}\cdot\dfrac{1}{2}\)
\(=\dfrac{12}{5}\)
B.1:
a) Với x = 1/2, y = -1/3, A= \(3\left(\frac{1}{2}\right)^3\left(-\frac{1}{3}\right)+6\left(\frac{1}{2}\right)^2\left(-\frac{1}{3}\right)^2+3.\frac{1}{2}.\left(-\frac{1}{3}\right)^3\)=\(\frac{-1}{8}+\frac{1}{6}+\frac{-1}{18}\)=\(\frac{-1}{72}\)
b)Với x = -1, y = 3, B=
\(\left(-1\right)^2.3^2+\left(-1\right).3+\left(-1\right)^3+3^3\)\(=9+\left(-3\right)+\left(-1\right)+27\)
\(=32\)
B.2:
\(P\left(-1\right)=\left(-1\right)^4+2.\left(-1\right)^2+1\)\(=1+2+1=4\)
\(P\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^4+2.\left(\frac{1}{2}\right)^2+1\)\(=\frac{1}{16}+\frac{1}{2}+1\)\(=\frac{25}{16}\)
\(Q\left(-2\right)=\left(-2\right)^4+4\left(-2\right)^3+2\left(-2\right)^2-4\left(-2\right)+1\)\(=16+\left(-32\right)+8-\left(-8\right)+1=1\)
\(Q\left(1\right)=1^4+4.1^3+2.1^2=1+4+2=7\)
Chúc cậu học tốt
=> 2x . 23 + 2x . 2 = 160
=> 2x (23 + 2) = 160
=> 2x = 16
=> 2x = 24
=> x = 4