A=ghi laị biểu thức 

A=(0,8*7+0,8*0,8)*(1,25*7-1,25*4/5)+31,64

A=[0,8*(7+0,8)]*[1,25*(7-4/5)]+31,64

A=(0,8*7,8)*(1,25*6,2)+31,+31,64

A=6,24*7,75+31,64

A=48,36+31,64

A=80

Lời giải:

Ta có:

\(\cos (\overrightarrow{a}, \overrightarrow{b})=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{2.6+(-3).4}{\sqrt{2^2+(-3)^2}.\sqrt{6^2+4^2}}\)

\(=0\)

\(\Rightarrow \angle (\overrightarrow{a}; \overrightarrow{b})=\frac{\pi}{2}\)

1.

\(\overrightarrow{AB}=\left(6;2\right)\) ; \(\overrightarrow{AC}=\left(1;-3\right)\) 

\(\Rightarrow\overrightarrow{AB}.\overrightarrow{AC}=6.1+2.\left(-3\right)=0\)

\(\Rightarrow AB\perp AC\Rightarrow\Delta ABC\) vuông tại A (ko phải tại B như đề bài ghi)

\(AB=\sqrt[]{6^2+2^2}=2\sqrt{10}\) ; \(AC=\sqrt{1^2+\left(-3\right)^2}=\sqrt{10}\)

\(S_{ABC}=\dfrac{1}{2}AB.AC=10\)

2.

Pt có 2 nghiệm cùng dấu khi:

\(\left\{{}\begin{matrix}\Delta'=16-\left(m+5\right)\ge0\\x_1x_2=m+5>0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\le11\\m>-5\end{matrix}\right.\)

\(\Rightarrow-5< m\le11\)

cos(\(\dfrac{3a}{2}\))*cos(\(\dfrac{a}{2}\))=\(\dfrac{1}{2}\left(cos\left(\dfrac{3a}{2}+\dfrac{a}{2}\right)+cos\left(\dfrac{3a}{2}-\dfrac{a}{2}\right)\right)\)=\(\dfrac{1}{2}\left(cos\left(2a\right)+cos\left(a\right)\right)\)=\(\dfrac{1}{2}\left(2cos^2a-1+cosa\right)\)=\(\dfrac{1}{2}\left(2\cdot\left(\dfrac{3}{4}\right)^2-1+\dfrac{3}{4}\right)=\dfrac{7}{16}\)

đề bài tính "A" :

\(\left\{{}\begin{matrix}\dfrac{x}{x^2-x+1}=a\\A=\dfrac{x^2}{x^4+x^2+1}\end{matrix}\right.\) \(\begin{matrix}\left(1\right)\\\\\left(2\right)\end{matrix}\)

\(x=0;a=0;A=0\)

\(x\ne0;\left(1\right)\Leftrightarrow\dfrac{1}{a}=\dfrac{x^2-x+1}{x}=x+\dfrac{1}{x}-1\)

\(\left(2\right)\Leftrightarrow\dfrac{1}{A}=\dfrac{x^4+x^2+1}{x^2}=x^2+\dfrac{1}{x^2}+1=\left(x+\dfrac{1}{x}\right)^2-1=\left(x+\dfrac{1}{x}-1\right)\left(x+\dfrac{1}{x}+1\right)\)

\(\dfrac{1}{A}=\dfrac{1}{a}\left(\dfrac{1}{a}+2\right)=\dfrac{2a+1}{a^2}\)

\(a=\dfrac{-1}{2}\Leftrightarrow\left(x^2+x+1\right)=0;voN_0\)

a khác -1/2 mọi x

\(A=\dfrac{a^2}{2a+1}\)

Điểm tổng hk2 là 8,8 - GDQP là 6,4 mà cả năm là 8,7 - GDQP là 6,6 thì cả năm là gì bạn ơi?

Bạn ơi trả lời giúp tui vói

\(\overrightarrow{AB}\cdot\overrightarrow{CB}=4\)

=>AB*CB*cosB=4

=>AB*CB*AB/BC=4

=>BA^2=4

=>AB=2

\(\overrightarrow{AC}\cdot\overrightarrow{BC}=9\)

=>AC*BC*cosC=9

=>AC*BC*AC/BC=9

=>AC=3

=>\(BC=\sqrt{13}\)