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a, Đặt \(2^x=t,t>0\)
Pt trở thành: \(t^2-10t+16=0\Leftrightarrow\left(t-2\right)\left(t-8\right)=0\Leftrightarrow\orbr{\begin{cases}t=2\\t=8\end{cases}\left(tm\right)}\)
Nếu t=2 => x=1
nếu t=8=> x=3
Vậy x=...
b, Đặt: \(2x^2-3x-1=t\)
pt trở thành: \(t^2-3\left(t-4\right)-16=0\Leftrightarrow t^2-3t-4=0\Leftrightarrow\left(t+1\right)\left(t-4\right)=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=4\end{cases}}\)
* Nếu t=-1 <=> \(2x^2-3x-1=-1\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
* Nếu t=4 <=> \(2x^2-3x-1=4\Leftrightarrow2x^2-3x-5=0\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}\)
Vậy x=...
â) \(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\left(5-x\right)\left(2+3x\right)=\left(2+3x\right)\left(2-3x\right)\)
\(5-x=2-3x\)
\(2x=-3\)
\(x=\frac{-3}{2}\)
Vậy ......
b) \(25-x^2=4x\left(5+x\right)\)
\(\left(5+x\right)\left(5-x\right)=4x\left(5+x\right)\)
\(5-x=4x\)
\(5x=5\)
x=1
Vậy......
a) \(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
<=> \(\left(5-x\right)\left(2+3x\right)+9x^2-4=0\)
<=> \(\left(5-x\right)\left(2+3x\right)+\left(3x-2\right)\left(3x+2\right)=0\)
<=> \(\left(2+3x\right)\left(3x-2+5-x\right)=0\)
<=> \(\left(2+3x\right)\left(2x+3\right)=0\)
<=> \(\orbr{\begin{cases}2x+3=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{3}{2}\\x=-\frac{2}{3}\end{cases}}\)
b) \(25-x^2=4x\left(5+x\right)\)
<=> \(25-x^2-4x\left(5+x\right)=0\)
<=> \(\left(5-x\right)\left(5+x\right)-4x\left(5+x\right)=0\)
<=> \(\left(5+x\right)\left(5-x-4x\right)=0\)
<=> \(\left(5+x\right)\left(5-5x\right)=0\)
<=> \(\orbr{\begin{cases}5+x=0\\5-5x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-5\\x=1\end{cases}}\)
a)(x-3)2 -x(x-2)=0
x=\(\frac{9}{4}\)
b)3x(2-x)+4(x-2) =0
x=2
c)(x-1)2=(49-1)16
x=5308417
d)x3-6x2+9x=0
x=0
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
Ta có ; (x - 3)2 - x(x - 2) = 0
<=> x2 - 6x + 9 - x2 + 2x = 0
<=> -4x + 9 = 0
=> -4x = -9
=> x = \(\frac{9}{4}\)
a) ( 4x - 1 ) (x - 3) - ( x - 3 ) ( 5x + 2 ) = 0
<=> (x - 3)(4x - 1 - 5x - 2) = 0
<=> (x - 3)(-x - 3) = 0
<=> x = 3 hoặc x = -3
b) ( x + 3 ) ( x - 5 ) + ( x + 3 ) ( 3x - 4) = 0
<=> (x + 3)(x - 5 + 3x - 4) = 0
<=> (x + 3)(4x - 9) = 0
<=> x = -3 hoặc x = 9/4
c) ( x + 6 ) ( 3x - 1 )+ x2 - 36 = 0
<=> 3x^2 + 17x - 6 + x^2 - 36 = 0
<=> 4x^2 + 17x - 42 = 0
<=> 4x^2 + 24x - 7x - 42 = 0
<=> 4x(x + 6) - 7(x + 6) = 0
<=> (4x - 7)(x + 6) = 0
<=> x = -6 hoặc x = 7/4
d) ( x + 4 ) ( 5x + 9 ) - x2 + 16 = 0
<=> 5x^2 + 29x + 36 - x^2 + 16 = 0
<=> 4x^2 + 29x + 52 = 0
<=> 4x^2 + 16x + 13x + 42 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> x = -13/4 và x = -4
a. ( x + 3 )( x - 3 ) = 16
⇔x2-9=16
⇔x2-16-9=0
⇔x2-25=0
⇔(x-5)(x+5)=0
⇔\(\left[{}\begin{matrix}x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)
a) ( x + 3 )( x - 3 ) = 16
\(\Leftrightarrow x^2-9=16\)
\(\Leftrightarrow x^2-9-16=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)
Vậy: \(x=-5;5\)
b) x2 - 9x + 20 = 0
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy: \(x=4;5\)
c) 2/x2+2x+1 - 5/x2-2x+1 = 3/1-x2
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)^2}-\dfrac{5}{\left(x-1\right)^2}=-\dfrac{3}{x^2-1}\)
\(\Leftrightarrow\dfrac{2}{\left(x+1\right)^2}-\dfrac{5}{\left(x-1\right)^2}=-\dfrac{3}{\left(x-1\right)\left(x+1\right)}\)
\(ĐKXĐ:x\ne\pm1\)
\(\Rightarrow2\left(x-1\right)^2-5\left(x+1\right)^2=-3\left(x^2-1\right)\)
d) | x + 4 | = 2x + 1
\(\Leftrightarrow\left[{}\begin{matrix}x+4=2x+1\\-\left(x+4\right)=2x+1\end{matrix}\right.\)khi \(\begin{matrix}x\ge4\\x< 4\end{matrix}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x=1-4\\-x-2x=1-\left(-4\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\-3x=5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(KTM\right)\\x=\dfrac{5}{3}\left(TM\right)\end{matrix}\right.\)
Vậy: \(x=\dfrac{5}{3}\)
a) \(\left(x+2\right)^2-\left(x-1\right)\left(x+1\right)=-1\)
\(\Leftrightarrow x^2+4x+4-x^2+1+1=0\)
\(\Leftrightarrow4x+6=0\)
\(\Leftrightarrow4x=-6\Leftrightarrow x=-\dfrac{3}{2}\)
Vậy ...
b) sửa đề: \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x\right)^2-4^2-\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-4-1\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy ...