Xin lỗi bạn,mk ms học đến phân tích đa thức thành nhân tử nhóm nhiều hạng tử,còn phần này mk ms học còn yếu lắm.

1. \(-10x^2+11x+6\)

\(=-10x^2+15x-4x+6\)

\(=-5x\left(2x-3\right)-2\left(2x-3\right)\)

\(=\left(-5x-2\right)\left(2x-3\right)\)

2.\(10x^2-4x-6\)

\(=2\left(5x^2-2x-3\right)\)

\(=2\left(5x^2+3x-5x-3\right)\)

\(=2\left[x\left(5x+3\right)-\left(5x+3\right)\right]\)

\(=2\left(x-1\right)\left(5x+3\right)\)

3. \(10x^2+7x-6\)

\(=10x^2+12x-5x-6\)

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

\(=\left(2x-1\right)\left(5x+6\right)\)

4. \(10x^2-14x-12\)

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

\(=2\left(5x^2+3x-10x-6\right)\)

\(=2\left[x\left(5x+3\right)-2\left(5x+3\right)\right]\)

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

a) Ta có: A = 0

=> x² + 2x - 3 = 0

=> x² + 3x - x - 3 = 0

=> x(x + 3) - (x + 3) = 0

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

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

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

Vậy ...

b) Ta có: B = 0

=> -3x² + 12x - 9 = 0

=> -3x² + 3x + 9x - 9 = 0

=> -3x(x - 1) + 9(x - 1) = 0

=> (-3x + 9)(x - 1) = 0

=> -3(x - 3)(x - 1) = 0

=> (x - 3)(x - 1) = 0

=> \(\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\)

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

Vậy ...

c) C = 0

=>  10x² - 7x - 3 = 0

=> 10x² - 10x + 3x - 3 = 0

=> 10x(x - 1) + 3(x - 1) = 0

=> (10x  + 3)(x - 1) = 0

=> \(\orbr{\begin{cases}10x+3=0\\x-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}10x=-3\\x=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

d) D = 0

=> -7x⁴ + 10x³ - 3x² = 0

=> x²(-7x² + 10x - 3) = 0

=> x²(-7x² + 7x + 3x - 3) = 0

=> x².[-7x(x - 1) + 3(x - 1)] = 0

=> x².(-7x + 3)(x - 1) = 0

=> x^2 = 0

-7x + 3 = 0

hoặc  x - 1 = 0

=> x=  0

-7x = -3

hoặc x = 1

=> x = 0

hoặc x = 3/7

hoặc x = 1

Vậy ...

a) \(A=x^2+2x-3\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)

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

b) \(B=-3x^2+12x-9\)

\(-3x^2+12x-9=0\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

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

c) \(C=10x^2-7x-3\)

\(10x^2-7x-3=0\)

\(\Leftrightarrow x\left(10x+3\right)-\left(10x+3\right)=0\)

\(\Leftrightarrow\left(10x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}10x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)

d) \(D=-7x^4+10x^3-3x^2\)

\(-7x^4+10x^3-3x^2=0\)

\(\Leftrightarrow-x^2\left(7x^2-10x+3\right)=0\)

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

\(\Leftrightarrow-x^2\left[x\left(7x-3\right)-\left(7x-3\right)\right]=0\)

\(\Leftrightarrow-x^2\left(7x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow x^2\left(7x-3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x^2=0\\7x-3=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{3}{7}\\x=1\end{cases}}\)(thay ngoặc nhọn bằng ngoặc vuông nhé, phần kl cũng thay luôn như thế nhé)

\(\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{7}\\x=1\end{cases}}\)

đề là gì vậy bạn

$2x^4-7x^3+10x^2-7x+2=0$

\(\Leftrightarrow x^2\left(2x^2-7x+10-\dfrac{7}{x}+\dfrac{2}{x^2}\right)=0\)

\(\Leftrightarrow2x^2-7x+10-\dfrac{7}{x}+\dfrac{2}{x^2}=0\)

\(\Leftrightarrow\left(2x^2+\dfrac{2}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+10=0\)

\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}\right)-7\left(x+\dfrac{1}{x}\right)+10=0\)

Đặt \(x+\dfrac{1}{x}\) là a ,thì \(x^2+\dfrac{1}{x^2}\) là a²-2 ta được

2(a²-2)-7a+10=0

⇔2a²-4-7a+10=0

⇔2a²-7a+6=0

⇔2a²-4a-3a+6=0

⇔(2a²-4a)-(3a-6)=0

⇔2a(a-2)-3(a-2)=0

⇔(a-2)(2a-3)=0

\(\Rightarrow\left[{}\begin{matrix}a-2=0\\2a-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=2\\a=\dfrac{3}{2}\end{matrix}\right.\)

ta được a=2 => \(x+\dfrac{1}{x}=2\) => x=1

a=\(x+\dfrac{1}{x}=\dfrac{3}{2}\)(vô nghiệm)

vậy S={1}

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

Mik ghi nhầm số 30 thành 20. Xin lỗi nhé!

đề???