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AH
Akai Haruma
Giáo viên
22 tháng 12 2022
Lời giải:
$2^{x+2}+2^{x+1}-2^x=40$
$2^x.2^2+2^x.2-2^x=40$
$2^x(2^2+2-1)=40$
$2^x.5=40$
$2^x=40:5=8=2^3$
$\Rightarrow x=3$
BN
10 tháng 4 2017
a) \(A\left(x\right)+B\left(x\right)=4x^5-2x^2-1\)
\(\Rightarrow B\left(x\right)=4x^5-2x^2-1-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=4x^5-2x^2-1-\left(2x^4-3x^3+\dfrac{1}{2}-4x\right)\)
\(B\left(x\right)=4x^5-2x^2-1-2x^4+3x^3-\dfrac{1}{2}+4x\)
Vậy \(B\left(x\right)=4x^5-2x^4+3x^3-2x^2+4x-\dfrac{3}{2}\)
b) \(A\left(x\right)-C\left(x\right)=2x^3\)
\(\Rightarrow C\left(x\right)=2x^3+A\left(x\right)\)
\(\Rightarrow C\left(x\right)=2x^3+2x^4-3x^3+\dfrac{1}{2}-4x\)
Vậy \(C\left(x\right)=2x^4-x^3-4x+\dfrac{1}{2}\)
5 tháng 3 2022
\(\Leftrightarrow4x\cdot2x-4x\cdot10x=400\)
\(\Leftrightarrow-32x^2=400\)(vô lý)
27 tháng 9 2019
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
4 tháng 10 2016
Bài 1:
\(\left(2x+1\right)^3=9\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)^3-9\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left[\left(2x+1\right)^2-9\right]=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1-3\right)\left(2x+1+3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-2\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+1=0\\2x-2=0\\2x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=1\\x=-2\end{array}\right.\)
Bài 2:
\(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\)
Vì: \(\left(2x-1\right)^2+\left(3-y\right)^2\ge0\)
=> \(\left(2x-1\right)^2+\left(3-y\right)^2+2017\ge2017\)
Dấu "=" xảy ra khi \(x=\frac{1}{2};y=3\)
Vậy GTNN của A là 2017 khi \(x=\frac{1}{2};y=3\)
4 tháng 10 2016
Bài 1:
(2x + 1)3 = 9.(2x + 1)
=> (2x + 1)3 - 9.(2x + 1) = 0
=> (2x + 1).[(2x + 1)2 - 9] = 0
=> (2x + 1).(2x + 1 - 3).(2x + 1 + 3) = 0
=> (2x + 1).(2x - 2).(2x + 4) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}2x+1=0\\2x-2=0\\2x+4=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}2x=-1\\2x=2\\2x=-4\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{-1}{2}\\x=1\\x=-2\end{array}\right.\)
Vậy \(x\in\left\{\frac{-1}{2};1;-2\right\}\)
Bài 2:
Có: \(\left(2x-1\right)^2\ge0;\left(3-y\right)^2\ge0\forall x;y\)
=> \(A=\left(2x-1\right)^2+\left(3-y\right)^2+2017\ge2017\)
Dấu "=" xảy ra khi và chỉ khi \(\begin{cases}\left(2x-1\right)^2=0\\\left(3-y\right)^2=0\end{cases}\)\(\Rightarrow\begin{cases}2x-1=0\\3-y=0\end{cases}\)\(\Rightarrow\begin{cases}2x=1\\y=3\end{cases}\)\(\Rightarrow\begin{cases}x=\frac{1}{2}\\y=3\end{cases}\)
Vậy GTNN của A là 2017 khi và chỉ khi \(x=\frac{1}{2};y=3\)
NM
1
18 tháng 7 2018
a/ Ta có :
\(\left|x-2\right|=\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|=\left|x\right|+\left|2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge\left|x+2-x\right|\)
\(\Leftrightarrow\left|x\right|+\left|x-2\right|\ge2\)
Dấu "=" xảy ra khi :
\(x\left(x-2\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-2\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x\ge2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x\le2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow2\le x\le2\Leftrightarrow x=2\)
Vậy \(x=2\)
b/ \(\left|2x-1\right|+\left|9-2x\right|\ge\left|2x-1+9-2x\right|\)
\(\Leftrightarrow\left|2x-1\right|+\left|9-2x\right|\ge8\)
Dấu "=" xảy ra khi :
\(\left(2x-1\right)\left(9-2x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-1\ge0\\9-2x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1\le0\\9-2x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\le\dfrac{9}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\ge\dfrac{9}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{1}{2}\le x\le\dfrac{9}{2}\)
Vậy ....
c/ Ta có : \(\left|3x-20\right|=\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|=\left|3x+7\right|+\left|20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge\left|3x+7+20-3x\right|\)
\(\Leftrightarrow\left|3x+7\right|+\left|3x-20\right|\ge27\)
Dấu "=" xảy ra khi :
\(\left(3x+7\right)\left(20-3x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x+7\ge0\\20-3x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}3x+7\le0\\20-3x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-\dfrac{7}{3}\\x\le\dfrac{20}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-\dfrac{7}{3}\\x\ge\dfrac{20}{3}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\dfrac{20}{3}\le x\le-\dfrac{7}{3}\)
Vậy...
d/ \(\left|10-x\right|+\left|x+30\right|\ge\left|10-x+x+30\right|\)
\(\Leftrightarrow\left|10-x\right|+\left|x+30\right|\ge40\)
Dấu "=" xảy ra khi :
\(\left(10-x\right)\left(x+30\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10-x\ge0\\x+30\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}10-x\le0\\x+30\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}10\ge x\\x\ge-30\end{matrix}\right.\\\left\{{}\begin{matrix}10\le x\\x\le-30\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow10\ge x\ge-30\)
Vậy...
2x+2+2x+1-2x = 40
2x.22+2x.2-2x = 40
2x(4+2-1) = 40
2x.5 = 40
2x = 8
x = 3
\(2^x.2^2+2^x.2-2^x=40\)
\(2^x\left(2^2+2-1\right)=40\)
\(2^x.5=40\)
\(2^x=8\)
\(x=3\)