a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

`P(x)=2x^5+2x+8-3x^2-2x^5+2x^3-4x^4 `

`= (2x^5 -2x^5) -4x^4 +2x^3 -3x^2 +2x +8`

`= -4x^4 +2x^3 -3x^2 +2x +8`

Bậc của `P(x)=10`

b: Ta có: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)

\(\Leftrightarrow\left(3x+4-2x-7\right)\left(3x+4+2x+7\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(5x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{5}\end{matrix}\right.\)

c: ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)

\(\Leftrightarrow\left(3x-6\right)^2-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(3x-6-x-2\right)\left(3x-6+x+2\right)=0\)

\(\Leftrightarrow\left(2x-8\right)\left(4x-4\right)=0\)

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

\(A=3\left(2x-3\right)\left(3x+2\right)-\left(2x+4\right)\left(4x-3\right)+9x\left(4-x\right)\)

\(=\left(6x-9\right)\left(3x+2\right)-8x^2+6x-16x+12+36x-9x^2\)

\(=18x^2+12x-27x-18-17x^2+26x+12\)

\(=x^2+11x-6\)

Để A = 0

\(\Leftrightarrow x^2+11x-6=0\)

\(\Leftrightarrow\left(x^2+11x+\frac{121}{4}\right)-\frac{145}{4}=0\)

\(\Leftrightarrow\left(x+\frac{11}{2}\right)^2-\left(\frac{\sqrt{145}}{2}\right)^2=0\)

\(\Leftrightarrow\left(x+\frac{11}{2}-\frac{\sqrt{145}}{2}\right)\left(x+\frac{11}{2}+\frac{\sqrt{145}}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{145}-11}{2}\\x=\frac{-\sqrt{145}-11}{2}\end{matrix}\right.\)

Vậy..................

\(5\left(3-2x\right)+5\left(x-4\right)=6-4x\\ \Leftrightarrow15-10x+5x-20=6-4x\\ \Leftrightarrow-10x+5x+4x=6-15+20\\ \Leftrightarrow-x=11\\ \Leftrightarrow x=-11\)

a)\(2x^4-6x^3+x^2+6x-3=0\)

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)

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⇔(2 x ) ²+2.2 x .1+1 ²=0 ... c ,(3 x −4) ²−14(3 x −4)(6+3 x )+49(3 x +6)=16 ... ⇔9 x ²−24 x +16−126 x ²−252 x +168 x +336+147 x +294=16.

https://olm.vn/hoi-dap/detail/192758180810.html

\(-2x\left(6x-2\right)+3x\left(4x-7\right)=8\)

<=> \(-12x^2+4x+12x^2-21x=8\)

<=> \(-17x=8\)

<=> \(x=-\frac{8}{17}\)

\(\left(7x-2\right)\left(2x+3\right)-\left(3x-5\right)\left(4x+6\right)=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30-2x^2+5=0\)

<=> \(19x+29=0\)

<=> \(19x=-29\)

<=> \(x=-\frac{29}{19}\)

Ý cuối mình k biết -22x2 là -22.2 hay -22x² nữa :)