2 x + 2X+1 + 2x+2 =56
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a) \(7x^2-16x=2x^3-56\)
\(\Leftrightarrow\)\(2x^3-7x^2+16x-56=0\)
\(\Leftrightarrow\)\(2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x-7\right)\left(x^2+8\right)=0\)
\(\Leftrightarrow\)\(2x-7=0\)
\(\Leftrightarrow\)\(x=3,5\)
Vậy...
b) \(x^7+x^3+2x^5+2x=0\)
\(\Leftrightarrow\)\(x.\left(x^6+x^2+2x^4+2\right)=0\)
\(\Leftrightarrow\)\(x\left(x^2+2\right)\left(x^4+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
Vậy...
c) \(\left(2x+1\right)x-5\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(2x\left(x+\frac{1}{2}\right)-5\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(\left(2x-5\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x-5=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2,5\\x=-0,5\end{cases}}\)
Vậy...
a: \(\Leftrightarrow2x^3-56-7x^2+16x=0\)
\(\Leftrightarrow2x\left(x^2+8\right)-7\left(x^2+8\right)=0\)
=>2x-7=0
hay x=7/2
b: \(\Leftrightarrow x^5\left(x^2+2\right)+x\left(x^2+2\right)=0\)
=>x(x2+2)(x4+1)=0
=>x=0
c: \(\Leftrightarrow2x^2+x-5x-\dfrac{5}{2}=0\)
\(\Leftrightarrow2x^2-4x-\dfrac{5}{2}=0\)
hay \(x\in\left\{\dfrac{5}{2};-\dfrac{1}{2}\right\}\)
\(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x+2^x.2+2^x.2^2=56\)
\(2^x.\left(1+2+4\right)=56\)
\(2^x.7=56\)
\(2^x=8\)
\(=>2^x=2^3\)
\(=>x=3\)
`2^x+2^x*2+2^x*2^2=56`
`2^x+2^x*2+2^x*4=56`
`2^x*(1+2+4)=56`
`2^x*7=56`
`2^x=56 \div 7`
`2^x=8`
`2^x=2^3`
`-> x=3`
56-(2x-1)^2=|9-49|
<=>56-(2x-1)^2=40
<=>(2x-1)^2=16
<=>2x-1=4
<=>x=5/2
Giải:
56-(2x-1)2=|32-72|
56-(2x-1)2=40
(2x-1)2=56-40
(2x-1)2=16
⇒(2x-1)2=42 hoặc (2x-1)2=(-4)2
2x-1=4 hoặc 2x-1=-4
x=5/2 hoặc x=-3/2
Chúc bạn học tốt!
\(2^x+2^{x+1}+2^{x+2}=56\)
\(2^x\left(1+2+2^2\right)=56\)
\(2^x\cdot7=56\)
\(2^x=8\)
\(x=3\)
Cho mk sửa lại ở câu c là 2^n+2 + 2^n+1 - 2^n = 56 nha!
a, \(\left(\frac{1}{2}+x\right)^2=\left(\frac{1}{2}\right)^2+2.\frac{1}{2}.x+x^2=\frac{1}{4}+x+x^2\)
\(\left(2x+1\right)^2=\left(2x\right)^2+2.2x.1+1^2=4x^2+4x+1\)
b, \(\left(2x+3y\right)^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
\(\left(0,01+xy\right)^2=\frac{1}{10000}+\frac{1}{50}xy+x^2y^2\)
c, \(\left(x+1\right)\left(x-1\right)=x^2-1\)
d, \(\left(x-2y\right)\left(x-2y\right)=\left(x-2y\right)^2=x^2-4xy+4y^2\)
\(56.64=\left(60-4\right)\left(60+4\right)=60^2-4^2\)
2^x .(1+2+4)
2^x .7=56
2^x=56:7
2^x=8
2^x=2^3
=> x=3
chúc bạn học tốt nha
ủng hộ mk với nha
\(2^x+2^{x+1}+2^{x+2}=56\)
\(\Leftrightarrow2^x+2^x.2+2^x.4=56\)
\(\Leftrightarrow2^x\left(1+2+4\right)=56\)
\(\Leftrightarrow2^x.7=56\)
\(\Rightarrow2^x=56:7\)
\(\Rightarrow2^x=8\)
\(\Rightarrow2^x=2^3\)
\(\Rightarrow x=3\)
Vậy \(x=3\)