Hình tự vẽ! a) Vì AH là đường cao của \(\Delta ABC\Rightarrow\widehat{AHB}=\widehat{AHC}=90^0\)

Xét \(\Delta ABC\) có \(\widehat{AHC}=90^0\Rightarrow AH^2+HC^2=AC^2\) ( ĐL Pytago ) 

\(\Rightarrow AC^2=12^2+16^2=144+256=400=20^2\Rightarrow AC=20\left(cm\right)\)

b) Xét \(\Delta ABC\) có \(\widehat{AHB}=90^0\Rightarrow AH^2+HB^2=AB^2\) ( ĐL Pytago ) 

\(\Rightarrow HB^2=AB^2-AH^2=13^2-12^2=169-144=25=5^2\Rightarrow HB=5\left(cm\right)\)

\(BC=HB+HC=16+5=21\left(cm\right)\)

( a + b ) ( a + b - c ) - ( a + b )²

=a² + ab - ac + ab + b² - bc - a² - 2ab - b²

=( a² - a² ) + ( b² - b² ) + ( ab +ab - 2ab ) - ac - bc

=-ac - bc

kikiykhii

x² +2x+2=0

x²+2x=-2

x(x+2)=-2

\(\Rightarrow\orbr{\begin{cases}x=-2\\x+2=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-4\end{cases}}.}\)

Vậy x=-2 hoặc x=-4

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

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

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

\(\Rightarrow\left[x\left(x+1\right)+\left(x+1\right)\right]+1=0\)

\(\Rightarrow\left(x+1\right)\left(x+1\right)+1=0\)

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

Ta có :

\(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1\forall x\)

\(\Rightarrow\left(x+1\right)^2+1>0\forall x\left(2\right)\)

Từ (1) và (2).

\(\Rightarrow\)Mâu thuẫn nhau.

Do đó \(x=\varnothing\)

Vậy \(x=\varnothing\)

Sửa lại đề như sau:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

BÀI LÀM:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1}\times2+\frac{1}{2}\times3+\frac{1}{3}\times4+\frac{1}{4}\times5+\frac{1}{5}\times6+\frac{1}{6}\times7\)

\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)\)

\(=1-\frac{1}{7}\)

\(=\frac{6}{7}\)

Trả lời:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)

\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)

\(=\frac{1}{1}-\frac{1}{7}\)

\(=\frac{7}{7}-\frac{1}{7}\)

\(=\frac{6}{7}\)