( 2x - 1)^2 = ( 2x -1)^20
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\(\left(\dfrac{1}{20}-\dfrac{1}{5}\right)\cdot2^{x+1}+2^{x+2}=\dfrac{-148}{5}\)
\(\Rightarrow-\dfrac{3}{20}\cdot2^{x+1}+2^{x+1}\cdot2=-\dfrac{148}{5}\)
\(\Rightarrow2^{x+1}\cdot\left(-\dfrac{3}{20}+2\right)=\dfrac{-148}{5}\)
\(\Rightarrow2^{x+1}\cdot\dfrac{37}{20}=\dfrac{-148}{5}\)
\(\Rightarrow2^{x+1}=\dfrac{-148}{5}:\dfrac{37}{20}\)
\(\Rightarrow2^{x+1}=-16\)
Xem lại bài
\(\left(\dfrac{1}{20}-\dfrac{1}{5}\right)\cdot2^{x+1}+2^{x+2}=-\dfrac{148}{5}\)
\(\Leftrightarrow-\dfrac{3}{20}\cdot2^{x+1}+2^{x+1}\cdot2=\dfrac{-148}{5}\)
\(\Leftrightarrow2^{x+1}\left(-\dfrac{3}{20}+2\right)=\dfrac{-148}{5}\)
\(\Leftrightarrow2^{x+1}\cdot\dfrac{37}{20}=\dfrac{-148}{5}\)
\(\Leftrightarrow2^{x+1}=\dfrac{-148}{5}:\dfrac{37}{20}\)
\(\Leftrightarrow2^{x+1}=-16\)
\(\Leftrightarrow2^{x+1}=-2^4\)(vô lí)
Xem lại đề bài!
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a. 5x.(12x+7)-3x.(20x-5)=-150
x=-3
b. ( 2x-1).(3-x)+(x+4).(2x-5)=20
x=43/10
c. 9x2-1+(3x-1)2=0
x=1/3
d. 3x.(x-2)-(3x+2).(x-1)=7
x=-5/2
e. (2x-1)2-(2x+5).(2x-5)=20
x=3/2
f. 4x2-5=4
x=3/2
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
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1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)
2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)
3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)
4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)
5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)
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( 2x - 1 )2 - ( 2x + 5 )( 2x - 5 ) = 20
\(=4x^2+1-4x-\left(4x^2-25\right)=20\)
\(=4x^2+1-4x-4x^2+25=20\)
\(=26-4x=20\)
\(4x=26-20=6\)
\(x=\frac{6}{4}=1,5\)
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a: \(\Leftrightarrow2x^2-6x+x-3=5x+4-2\left(x^2-4\right)\)
\(\Leftrightarrow2x^2-5x-3=5x+4-2x^2+8\)
\(\Leftrightarrow4x^2-10x-9=0\)
\(\text{Δ}=\left(-10\right)^2-4\cdot4\cdot\left(-9\right)=100+144=244>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{10-\sqrt{244}}{8}=\dfrac{5-\sqrt{61}}{4}\\x_2=\dfrac{5+\sqrt{61}}{4}\end{matrix}\right.\)
b: \(\Leftrightarrow3x-4x^2-4x-1=4x^2-\left(2x^2+2x-x-1\right)\)
\(\Leftrightarrow-4x^2-x-1-4x^2+2x^2+x-1=0\)
\(\Leftrightarrow-6x^2-2=0\)
hay \(x\in\varnothing\)
c: \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{x^2+2x-3}\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)=x^2+2x-3-4\)
\(\Leftrightarrow3x^2+9x-x-3-2x^2+2x-5x+5=x^2+2x-7\)
=>5x+2=2x-7
=>3x=-9
hay x=-3(loại)
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a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\)
\(=10x-6x^3-60-10x\)
\(=\) \(-6x^3-60\)
a) \(2x\left(5-3x^2\right)-10\left(6+x\right)\\ =2x.5-2x.3x^2-10.6-10.x\\ =10x-6x^3-60-10x\)
b) \(3\left(-x+2\right)-6\left(1-x+5x^{20}\right)\\ =-3.x+3.2-6.1+6.x-5.5x^{20}\\ =-3x+6-6+6x-25x^{20}=25x^{20}+3x\)
c) \(7x\left(2-5x^2+\dfrac{1}{2}x^3\right)-14x\left(1-2x^2\right)\\ =7x.2-7x.5x^2+7x.\dfrac{1}{2}x^3-14x.1+14x.2x^2\\ =14x-25x^3+\dfrac{7}{2}x^4-14x+28x^3=3x^2+\dfrac{7}{2}x^4\)
x = 1
x = \(\frac{1}{2}\)= 0,5
x = 0