giúp iem vs
√(4x^2-12x+9)=7
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3: Ta có: \(\sqrt{4x+1}=x+1\)
\(\Leftrightarrow x^2+2x+1=4x+1\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)
4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)
\(\Leftrightarrow3\sqrt{x-1}=15\)
\(\Leftrightarrow x-1=25\)
hay x=26
5: Ta có: \(\sqrt{4x^2-12x+9}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a: =(x-y)^2+2(x-y)
=(x-y)(x-y+2)
c: =(x-3)(x+3)+(x-3)^2
=(x-3)(x+3+x-3)
=2x(x-3)
d: =(x+3)(x^2-3x+9)-4x(x+3)
=(x+3)(x^2-7x+9)
e: =(x^2-8x+7)(x^2-8x+15)-20
=(x^2-8x)^2+22(x^2-8x)+85
=(x^2-8x+17)(x^2-8x+5)
a, ĐKXĐ: \(x\le2\)
\(\sqrt{4-2x}=5\\ \Leftrightarrow4-2x=25\\ \Leftrightarrow2x=-21\\ \Leftrightarrow x=-10,5\left(tm\right)\)
b, ĐKXĐ: \(x\ge-1\)
\(\sqrt{25\left(x+1\right)}+\sqrt{9x+9}=16\\ \Leftrightarrow5\sqrt{x+1}+\sqrt{9\left(x+1\right)}=16\\ \Leftrightarrow5\sqrt{x+1}+3\sqrt{x+1}=16\\ \Leftrightarrow8\sqrt{x+1}=16\\ \Leftrightarrow\sqrt{x+1}=2\\ \Leftrightarrow x+1=4\\ \Leftrightarrow x=3\)
c, \(\sqrt{4x^2+12x+9}=4\Leftrightarrow4x^2+12x+9=16\\ \Leftrightarrow4x^2+12x-7=0\\ \Leftrightarrow\left(4x^2-2x\right)+\left(14x-7\right)=0\\ \Leftrightarrow2x\left(2x-1\right)+7\left(2x-1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
a: \(\Leftrightarrow4-2x=25\)
hay \(x=-\dfrac{21}{2}\)
c: \(\Leftrightarrow\left|2x+3\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=4\\2x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
3x(12x-4)-(4x-3)(9x+4) = 9
36x2 -12x-(36x2 -16x-27x-12) = 9
36x2 -12x-36x2 -16x+27x+12 = 9
-x = 9-12
-x = -3
x= -3 : -1
x= 3
vậy x= 3
\(\Leftrightarrow\sqrt{\left(2x-3\right)^2}=7\\ \Leftrightarrow\left|2x-3\right|=7\\ \Leftrightarrow\left[{}\begin{matrix}2x-3=7\\3-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
bài ni là giải phương trình nhoa