2x(x+7)=-16
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Đề:........
<=> x2. (2x + 7) - 16. (2x + 7) = 0
<=> (2x + 7). (x2 - 16) = 0
<=> (2x+ 7). (x - 4). (x + 4) = 0
=> \(\hept{\begin{cases}2x+7=0\\x-4=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=-7\\x=4\\x=-4\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{-7}{2}\\x=4\\x=-4\end{cases}}\)
Vậy...........
\(x^2\left(2x+7\right)=16\left(2x+7\right)\)
\(x^2\left(2x+7\right)-16\left(2x+7\right)=0\)
\(\left(2x+7\right)\left(x^2-16\right)=0\)
\(\left(2x+7\right)\left(x+4\right)\left(x-4\right)=0\)
\(\hept{\begin{cases}x=-\frac{7}{2}\\x=-4\\x=4\end{cases}}\)
\(\left(3x-2\right)\left(x^2+16\right)=\left(2x-7\right)\left(x^2+16\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2+16\right)-\left(2x-7\right)\left(x^2+16\right)=0\)
\(\Leftrightarrow\left(x^2+16\right)\left(3x-2-2x+7\right)=0\)
\(\Leftrightarrow\left(x^2+16\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+16=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{-16}\\x=-5\end{matrix}\right.\)
ta có:
\(\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)
\(\Leftrightarrow\left(16-2x+x^2-9+2x-x^2\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)
\(\Leftrightarrow7=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)
\(\Leftrightarrow\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)
Ta có:
\(\left(\sqrt{16-2x+x^2}+\sqrt{9-2x+x^2}\right)\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)=7\)
\(\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)
\(\Leftrightarrow\left(16-2x+x^2-9+2x-x^2\right)=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)
\(\Leftrightarrow7=7\left(\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}\right)\)
\(\Leftrightarrow\sqrt{16-2x+x^2}-\sqrt{9-2x+x^2}=1\)
Ủng hộ nha
a: =>3|x-4|=16-2x
TH1: x>=4
=>3x-12=16-2x
=>5x=28
=>x=28/5(nhận)
TH2: x<4
=>12-3x=16-2x
=>-x=4
=>x=-4(nhận)
b: =>|2x-5|=7+4x+4=4x+11
TH1: x>=5/2
=>4x+11=2x-5
=>2x=-16
=>x=-8(loại)
TH2: x<5/2
=>4x+1=5-2x
=>6x=4
=>x=2/3(nhận)
x^2 -2x = 24
=> x^2 - 2x - 24=0
=>x^2 -8x+6x - 24 = 0
=> ( x^2- 8x)+( 6x-24) = 0
=> x(x-8) + 6(x-8) = 0
=> (x+6)(x-8)=0
=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)
6x=-16+2x
=>6x-2x=-16
=>4x=-16
=>x=-4
Vậy x=4.
2(x-7)=-16
=>2x-14=-16
=>2x=-16+14
=>2x=-2
=>x=-1
Vậy x=-1.
3x+14=-16
3x=-16-14
3x=-30
x=-30:3
x=-10
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K MK NHA