nghiệm = 2

\(7\left(x^2-5x+6\right)\)

\(7\left(x^2-3x-2x+6\right)\)

\(7\left(x-3\right)\left(x-2\right)\)

3TH

TH1 : 7=0  ( vô nghiệm )

TH2 x=3 

TH3 x=2

Ai làm giúp mình với

\(7x^2-35x+42=0\)

\(\Leftrightarrow7\left(x^2-5x+6\right)=0\)

\(\Leftrightarrow7\left(x^2-2x-3x+6\right)\)

\(\Leftrightarrow7\left[x\left(x-2\right)-3\left(x-2\right)\right]\)

\(\Leftrightarrow7\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}7=0\\x-3=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=3\\x=2\end{matrix}\right.\)

Vậy x = 3 hoặc x = 2

thiếu quế bênh trái nha bn

xửa đề : tìm nghiệm của đa thức \(7x^2-35x+42=0\)

\(7x^2-35x+42\Leftrightarrow7x^2-14x-21x+42\)

\(\Leftrightarrow7x\left(x-2\right)-21\left(x-2\right)=0\Leftrightarrow\left(7x-21\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}7x-21=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=21\\x=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

vậy \(x=3;x=2\)

\(7x^2-35x+42=7x^2-14x-21x+42\)

\(=7x\left(x-2\right)-21\left(x-2\right)=\left(7x-21\right)\left(x-2\right)\)

đa thức có nghiệm <=> \(\left(7x-21\right)\left(x-2\right)=0\Leftrightarrow\int^{7x-21=0\Rightarrow x=3}_{x-2=0\Rightarrow x=2}\)

Vậy..........................

Bài 3:

a)

Chúc em học tốt!

a: Đặt x^2-20x-300=0

=>x^2-30x+10x-300=0

=>(x-30)(x+10)=0

=>x=30 hoặc x=-10

b: Đặt 2x^2+35x+33=0

=>(x+1)(2x+33)=0

=>x=-33/2 hoặc x=-1

a) \(f\left(x\right)=x^2+7x-8=0\)

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

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

\(\Leftrightarrow f\left(x\right)=x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow f\left(x\right)=\left(x-1\right)\left(x+8\right)=0\)

\(\Rightarrow x-1=0\) hoặc  \(x+8=0\)

Nếu \(x-1=0\Rightarrow x=1\) 

Nếu  \(x+8=0\Rightarrow x=-8\)

Vậy đa thức f(x) có nghiệm là 1 và -8

b) \(k\left(x\right)=5x^2+9x+4=0\)

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

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

\(\Leftrightarrow k\left(x\right)=5x\left(x+1\right)+4\left(x+1\right)=0\)

\(\Leftrightarrow k\left(x\right)=\left(x+1\right)\left(5x+4\right)=0\)

\(\Rightarrow x+1=0\) hoặc \(5x+4=0\)

Nếu \(x+1=0\Rightarrow x=-1\)

Nếu \(5x+4=0\Rightarrow x=-\frac{4}{5}\)

Vậy đa thức k(x) có nghiệm là -1 và -4/5

3x² + x - 10 = 0

<=> (x + 2)(3x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

2x² - x - 15 = 0

<=> (x - 3)(2x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

7x² + 23x - 20 = 0

<=> (7x - 5)(x + 4) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{7}\\x=-4\end{matrix}\right.\)

6x² - 35x + 25 = 0

<=> (x - 5)(6x - 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{6}\end{matrix}\right.\)

6x² - x - 2 = 0

<=> (3x - 2)(2x + 1) = 0

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

8x² + 6x - 35 = 0

<=> (4x - 7)(2x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{5}{2}\end{matrix}\right.\)