\(\left(8x+5\right)\left(8x+7\right)\left(8x+6\right)^2=72\)

Đặt \(8x+5=t\left(t\ge0\right)\)

\(t\left(t+2\right)\left(t+1\right)^2-72=0\)

\(\Leftrightarrow t\left(t+1\right)\left(t+2\right)\left(t+1\right)-72=0\)

\(\Leftrightarrow\left(t^2+t\right)\left(t^2+3t+2\right)-72=0\)

\(\Leftrightarrow t^4+3t^3+2t^2+t^3+3t^2+2t-72=0\)

\(\Leftrightarrow t^4+4t^3+5t^2+2t-72=0\)

\(\Leftrightarrow\left(t^2+2t+9\ne0\right)\left(t+4\right)\left(t-2\right)=0\Leftrightarrow t=-4;2\)

hay \(8x+5=-4\Leftrightarrow x=-\frac{9}{8}\)( trường hợp 1 ) 

\(8x+5=2\Leftrightarrow x=-\frac{3}{8}\)( trưởng hợp 2 ) 

Vậy tập nghiệm của phương trình là S = { -9/8 ; -3/8 }

\(\left(8x+5\right)\cdot\left(8x+7\right)\cdot\left(8x+6\right)^2=72\)

Đặt \(t=8x+6\)

\(Pt\Leftrightarrow\left(t-1\right)\left(t+1\right)t^2-72=0\)

\(\Leftrightarrow\left(t^2-1\right)t^2-72=0\Leftrightarrow t^4-t^2-72=0\)

\(\Leftrightarrow\left(t^2-9\right)\left(t^2+8\right)=0\Leftrightarrow\orbr{\begin{cases}t^2=9\\t^2=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}t=3\\t=-3\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}8x+6=3\\8x+6=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{8}\\x=-\frac{9}{8}\end{cases}}}\)

Vậy....

( 8x + 5 )( 8x + 7 )( 8x + 6 )² = 72

<=> ( 64x² + 96x + 35 )( 64x² + 96x + 36 ) - 72 = 0

Đặt t = 64x² + 96x + 35

pt <=> t( t + 1 ) - 72 = 0

<=> t² + t - 72 = 0

<=> t² - 8t + 9t - 72 = 0

<=> t( t - 8 ) + 9( t - 8 ) = 0

<=> ( t - 8 )( t + 9 ) = 0

<=> ( 64x² + 96x + 35 - 8 )( 64x² + 96x + 35 + 9 ) = 0

<=> ( 64x² + 96x + 27 )( 64x² + 96x + 44 ) = 0

<=> 4( 64x² + 24x + 72x + 27 )( 16x² + 24x + 11 ) = 0

<=> 4[ 8x( 8x + 3 ) + 9( 8x + 3 ) ]( 16x² + 24x + 11 ) = 0

<=> 4( 8x + 3 )( 8x + 9 )( 16x² + 24x + 11 ) = 0

<=> 8x + 3 = 0 hoặc 8x + 9 = 0

[ do 16x² + 24x + 11 = ( 16x² + 24x + 9 ) + 2 = ( 4x + 3 )² + 2 ≥ 2 ∀ x ]

<=> x = -3/8 hoặc x = -9/8

Vậy phương trình có tập nghiệm S = { -3/8 ; -9/8 }

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) 

\(P=\left(x^{14}-9x^{13}\right)-\left(x^{13}-9x^{12}\right)+\left(x^{12}-9x^{11}\right)-...+\left(x^2-9x\right)-\left(x-9\right)+1\)

\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+x^{11}\left(x-9\right)+...+x\left(x-9\right)-\left(x-9\right)+1\)

\(P\left(9\right)=1\)

b)

\(Q=\left(x^{15}-7x^{14}\right)-\left(x^{14}-7x^{13}\right)+\left(x^{13}-7x^{12}\right)-...-\left(x^2-7x\right)+\left(x-7\right)+2\)

\(=x^{14}\left(x-7\right)-x^{13}\left(x-7\right)+x^{12}\left(x-7\right)-...-x\left(x-7\right)+\left(x-7\right)+2\)

\(Q\left(7\right)=2\)

x bằng bao nhiêu

Đặt \(A=x^{13}-\left(8x^{12}-8x^{11}+8x^{10}-8x^9+.....+8x^2-8x^1\right)+8\)

Đặt \(B=8x^{12}-8x^{11}+8x^{10}-....+8x^2-8x^1\)

\(B=8.\left(x^{12}-x^{11}+x^{10}-x^9+....+x^2-x^1\right)\)

Đặt \(C=x^{12}-x^{11}+x^{10}-x^9+...+x^2-x\)

Suy ra \(C.x=x^{13}-x^{12}+x^{11}-x^{10}+.....+x^3-x^2\)

Nên \(C.x-C=x^{13}-x\)hay \(C.\left(x-1\right)=x^{13}-x\)

Khi đó \(C=\frac{x^{13}-x}{x-1}\)nên\(B=8.\frac{x^{13}-x}{x-1}\)

Từ đó tính tương tự nha , cách làm thì có thể sai những em vẫn cố gắng giúp , ai có cách hay hơn thì giải nhé 

chả hiểu gì

chả hiểu gì

Câu 5: B

Câu 6: 

a: ĐKXĐ: \(x-2\ne0\)

=>\(x\ne2\)

b: ĐKXĐ: \(x+1\ne0\)

=>\(x\ne-1\)

8:

\(A=\dfrac{x^2+4}{3x^2-6x}+\dfrac{5x+2}{3x}-\dfrac{4x}{3x^2-6x}\)

\(=\dfrac{x^2+4-4x}{3x\left(x-2\right)}+\dfrac{5x+2}{3x}\)

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

\(=\dfrac{x-2+5x+2}{3x}=\dfrac{6x}{3x}=2\)

7: 

\(\dfrac{8x^3yz}{24xy^2}\)

\(=\dfrac{8xy\cdot x^2z}{8xy\cdot3y}\)

\(=\dfrac{x^2z}{3y}\)

B = x15 - 8x14 + 8x13 - 8x2 + ... - 8x2 + 8x - 5

B = x^15 - 7x^14 -x^14+7x^13+x^13-7x^12-...-x^2+7x+x-5

B = x^14(x-7) - x^14(x-7) +...+x^2(x-7)-x(x-7)+x-5

B = 7-5=2

`B = x^15 - 7x^14 - x^14 + 7x^13 + x^13 - .... +7x + x - 7 + 2`

`<=> x^14(x-7) - x^13(x-7) + ... + x - 7 + 2`

`<=> (x^14-x^13 + ... + 1)(x-7) + 2`

Thay `x = 7 <=> (x^14 - x^13 + ... + 1) xx 0 + 2 = 2`.

B = \(x^{15}-8x^{14}+8x^{13}+8x^2+...-8x^2+8x-5\)

B = \(x^{15}-7x^{14}+x^{14}+7x^{13}+x^{13}+7x^{12}-...-x^2+7x+x-5\)

B = \(x^{14}\left(x-7\right)-x^{14}\left(x-7\right)+...+x^2\left(x-7\right)-x\left(x-7\right)-5\)

B = \(7-5=2\)