Question

[vyanh04]

Ai giả hộ mk mí câu này đc k ạ?

Answer 1

[BW_P&A]

(Bài 8)*:\(\left(x-4\right)\left(x-5\right)\left(x-6\right)\left(x-7\right)=1680\)

\(\Leftrightarrow\left(x-4\right)\left(x-7\right)\left(x-5\right)\left(x-6\right)=1680\)

\(\Leftrightarrow\left(x^2-7x-4x+28\right)\left(x^2-6x-5x+30\right)=1680\)

\(\Leftrightarrow\left(x^2-11x+28\right)\left(x^2-11x+30\right)=1680\)

Đặt: \(t=x^2-11x+28\)

\(\Leftrightarrow t\left(t+2\right)=1680\)

\(\Leftrightarrow t^2+2t+1=1680+1\)( cả 2 vế cộng cho 1)

\(\Leftrightarrow\left(t+1\right)^2=1681=41^2\)

\(\Leftrightarrow t+1=41\)

\(\Leftrightarrow t=40\)

\(\Leftrightarrow x^2-11x+28=40\)

\(\Leftrightarrow x^2-11x+28-40=x^2-11x-12=0\)

\(\Leftrightarrow x^2-x+12x-12=0\)

\(\Leftrightarrow x\left(x-1\right)+12\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+12\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x-1=0\\x+12=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=1\\x=-12\end{matrix}\right.\)

Vậy: Tập nghiệm của phương trình là: \(S=\left\{-12;1\right\}\)

Answer 2

[diem ngo]

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

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

2. <=> x²(x²+x-12)=0

<=> \(\left[\begin{matrix}x=0\\x=-4\\x=3\end{matrix}\right.\)

3. (x⁴-x)-3x²(x-1) =0

<=> x(x-1)(x²+x+1)-3x²(x-1)=0

<=> x(x-1)(x²+x+1-3x)=0

<=> \(\left[\begin{matrix}x=0\\x=1\end{matrix}\right.\)

4. <=> x²(x-4)-(x-4) =0

<=> (x-4)(x²-1)=0

<=> \(\left[\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)