\(x^2-11x-26=0\)

\(x^2-13x+2x-26=0\)

\(x.\left(x-13\right)+2.\left(x-13\right)=0\)

\(\left(x+2\right).\left(x-13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-13=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-2\\x=13\end{cases}}\)

vậy...

P/S: lớp 7 sai sót mong thông cảm

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

\(2x^2+8x-x-4=0\)

\(2x.\left(x+4\right)-\left(x+4\right)=0\)

\(\left(2x-1\right).\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=1\\x=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=-4\end{cases}}\)

vậy ...

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

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

\(\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)

vậy ...

nghiệm của đa thức là -1

Tìm nghiệm bạn chỉ cần thay và thử giá trị là ra thôi. nhớ kb nhá

chúc bạn học tốt~

P(x) = 8x² + 11x + 3 

= 8x² + 8x + 3x + 3

= 8x ( x + 1 ) + 3(x+1)

= ( 8x + 3 ) ( x+1 )

Cho p(x) = 0

.\(=>\orbr{\begin{cases}x+1=0\\8x+3=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=-1\\x=\frac{-3}{8}\end{cases}}\)

Để P(x) = 0

=> 8x^2 + 11x +3 =0

8x^2 + 8x + 3x +3 = 0

8x.( x+1) + 3.(x+1) =0

(x+1) .( 8x+3) =0

=> x+1 = 0 => x = -1

=> 8x+3 = 0 => 8x = -3 => x = -3/8

KL: x = -1 ; x = -3/8 là nghiệm của đa thức P(x)

a ) \(x^2-11x-26=0\)

\(\Leftrightarrow x^2-13x+2x-26=0\)

\(\Leftrightarrow x\left(x-13\right)+2\left(x-13\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=13\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-2\\x=13\end{matrix}\right.\)

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

\(\Leftrightarrow2\left(x^2+\dfrac{7}{2}x-2\right)=0\)

\(\Leftrightarrow x^2+\dfrac{7}{2}x-2=0\)

\(\Leftrightarrow x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{81}{16}=0\)

\(\Leftrightarrow\left(x+\dfrac{7}{4}\right)^2=\dfrac{81}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{9}{4}\\x+\dfrac{7}{4}=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-4\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-4\end{matrix}\right.\)

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

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

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

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

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

Vậy \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Áp dụng TCDTSBN ta có:

\(\dfrac{3x}{2}=\dfrac{2y}{5}=\dfrac{6x-6y}{4-15}=-\dfrac{108}{11}\)

\(\dfrac{3x}{2}=-\dfrac{108}{11}\Rightarrow x=-\dfrac{72}{11}\\ \dfrac{2y}{5}=-\dfrac{108}{11}\Rightarrow y=-\dfrac{270}{11}\)

      \(\frac{11x-2}{7x-5}=\frac{11}{8}\)

\(\Leftrightarrow8\left(11x-2\right)=11\left(7x-5\right)\)

\(\Leftrightarrow88x-16=77x-55\)

\(\Leftrightarrow88x-77x=-55+16\)

\(\Leftrightarrow11x=-39\)

\(\Leftrightarrow x=\frac{-39}{11}\)

\(\frac{12x-2}{7x-8}=\frac{11}{8}\)

\(\Rightarrow8\left(12x-2\right)=11\left(7x-8\right)\)

\(\Rightarrow96x-16=77x-88\)

\(\Rightarrow96x-77x=-88+16\)

\(\Rightarrow19x=-72\)

\(\Rightarrow x=-\frac{72}{19}\)

8.(11x-2)=11.(7x-5)

88x-16=77x-55

88x-77x=16-55

11x=-39

x=-39:11

x=-39/11

nhớ k cho mình nha

a: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{2}\\x=1\end{matrix}\right.\)