\(D=50^2-49.51\)

\(\Leftrightarrow D=50^2-\left(50-1\right)\left(50+1\right)\)

\(\Leftrightarrow D=50^2-50^2+1=1\)

\(C=39^2+78.61+61^2\)

\(\Leftrightarrow C=39^2+2.39.61+61^2\)

\(\Leftrightarrow C=\left(39+61\right)^2=100^2=10000\)

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

\(\Leftrightarrow2x\left(3x+1\right).2x=0\)

\(\Leftrightarrow4x^2\left(3x+1\right)=0\)

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

Ta có: \(\left(x+1\right)\left(6x^2+2x\right)+\left(x-1\right)\left(6x^2+2x\right)=0\)

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

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

\(\Leftrightarrow2x.\left(3x+1\right).2x=0\)

\(\Leftrightarrow4x^2\left(3x+1\right)=0\)

\(\Leftrightarrow4x^2=0\text{ hoặc }3x+1=0\)

\(\Leftrightarrow x=0\text{ hoặc }x=\frac{-1}{3}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

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

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

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

\(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15=0\)

\(\Leftrightarrow\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15=0\)

Đặt \(x^2+8x+11=y\Rightarrow x^2+8x+7=y-4;x^2+8x+15=y+4\)

Khi đó:

\(pt\Leftrightarrow\left(y-4\right)\left(y+4\right)+15=0\)

\(\Leftrightarrow y^2-1=0\)

\(\Leftrightarrow y=1;y=-1\)

Nếu \(y=1\Rightarrow x^2+8x+11=1\)

\(\Rightarrow x^2+8x+10=0\)

\(\Rightarrow-\left(6-x^2-8x-16\right)=0\)

\(\Rightarrow-\left[6-\left(x+4\right)^2\right]=0\)

\(\Rightarrow-\left(\sqrt{6}-x-4\right)\left(\sqrt{6}+x+4\right)=0\)

\(\Rightarrow x=-4-\sqrt{6};x=\sqrt{6}-4\)

Nếu \(y=-1\),ta có:

\(x^2+8x+11=-1\)

\(\Rightarrow x^2+8x+12=0\)

\(\Rightarrow x^2+2x+6x+12=0\)

\(\Rightarrow x\left(x+2\right)+6\left(x+2\right)=0\)

\(\Rightarrow\left(x+2\right)\left(x+6\right)=0\)

\(\Rightarrow x=-2;x=-6\)

Vậy \(x=-2;x=-6;x=-4-\sqrt{6};x=\sqrt{6}-4\)

Hình bạn tự vẽ nhé...

a)

Xét tam giác BAH và tam giác ABC , có :

A^ = H^ = 90O

B^ : góc chung

=> tam giác HAB ~ tam giác ACB ( g.g)

c) 

 ADĐL pitago vào tam giác vuông ABC , có :

AB² + AC² = BC²

=> 12² + 16⁶ = BC²

=> BC² = 400

=> BC = 20 cm

Vì tam giác ACB ~ tam giác HAB , nên ta có :

 AH/AC= AB/BC

=> AH/16=12/20

=> AH = 9,6 cm.

Vì \(x=2017\Rightarrow x+1=2018\)

Thay \(x+1=2018\)vào biểu thức A ta được :

\(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+\left(x+1\right)\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1\)

\(=1\)

Tại x=2017 thì 2018 = x + 1 

Khí đó \(A=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-...-\left(x+1\right)x+x+1\)

\(=x^{10}-x^{10}-x^9+x^9+x^8-...-x^2-x+x+1\)

\(=1\)