lớp 8 có pt bậc 2 ak??

\(m,x^2+6x-16=0\)

\(\Leftrightarrow x^2-2x+8x-16=0\)

\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)

\(n,2x^2+5x-3=0\)

\(\Leftrightarrow2x^2-x+6x-3=0\)

\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

\(k,x\left(2x-7\right)-4x+14=0\)

\(\Leftrightarrow2x^2-4x-7x+14=0\)

\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)

b) Ta có: \(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)

\(=x^3+2x^2-7x-14-\left(2x^2-28x-x+14\right)+x^3-2x^2-22x+35\)

\(=2x^3-29x+21-2x^2+29x-14\)

\(=2x^3-2x^2+7\)

1.

Đặt \(x-2=t\ne0\Rightarrow x=t+2\)

\(B=\dfrac{4\left(t+2\right)^2-6\left(t+2\right)+1}{t^2}=\dfrac{4t^2+10t+5}{t^2}=\dfrac{5}{t^2}+\dfrac{2}{t}+4=5\left(\dfrac{1}{t}+\dfrac{1}{5}\right)^2+\dfrac{19}{5}\ge\dfrac{19}{5}\)

\(B_{min}=\dfrac{19}{5}\) khi \(t=-5\) hay \(x=-3\)

2.

Đặt \(x-1=t\ne0\Rightarrow x=t+1\)

\(C=\dfrac{\left(t+1\right)^2+4\left(t+1\right)-14}{t^2}=\dfrac{t^2+6t-9}{t^2}=-\dfrac{9}{t^2}+\dfrac{6}{t}+1=-\left(\dfrac{3}{t}-1\right)^2+2\le2\)

\(C_{max}=2\) khi \(t=3\) hay \(x=4\)

a: Đặt |x|=a

Pt trở thành \(3a^2-14a-5=0\)

=>(a-5)(3a+1)=0

=>a=5(nhận) hoặc a=-1/3(loại)

=>x=-5 hoặc x=5

c: \(\left|x+2\right|-2x+1=x^2+2x+3\)

\(\Leftrightarrow\left|x+2\right|=x^2+4x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+4x+2>=0\\\left(x^2+4x+2-x-2\right)\left(x^2+4x+2+x+2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+5x+4\right)=0\)

hay \(x\in\left\{0;-3;-1;-4\right\}\)

\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\)

\(< =>\left(1-x\right)\left(5x+3+3x-7\right)=0\)

\(< =>\left(1-x\right)\left(8x-4\right)=0\)

\(< =>\orbr{\begin{cases}1-x=0\\8x-4=0\end{cases}< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)

\(\left(x-2\right)\left(x+1\right)=x^2-4\)

\(< =>\left(x-2\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\)

\(< =>\left(x-2\right)\left(x+1-x-2\right)=0\)

\(< =>-1\left(x-2\right)=0\)

\(< =>2-x=0< =>x=2\)

\(2x^3+3x^2-32x=48\)

\(< =>x^2\left(2x+3\right)-16\left(2x+3\right)=0\)

\(< =>\left(x^2-16\right)\left(2x+3\right)=0\)

\(< =>\left(x-4\right)\left(x+4\right)\left(2x+3\right)=0\)

\(< =>\hept{\begin{cases}x=4\\x=-4\\x=-\frac{3}{2}\end{cases}}\)

Bài 6:

c: \(9x^2+6x+1=\left(3x+1\right)^2\)

d: \(4x^2-9=\left(2x-3\right)\left(2x+3\right)\)

e: \(x^3+27=\left(x+3\right)\left(x^2-3x+9\right)\)

\(a,2\left(x-1\right)+3=x+2\)

\(\Leftrightarrow2x-2+3=x+2\)

\(\Leftrightarrow2x-x=2+2-3\)

\(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

\(b,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)

\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(5+x\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(5x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-5;2\right\}\)

Đáp án cần chọn là: C