\(A=\frac{12\cdot25+4\cdot175\cdot3}{36\cdot\left(101-1\right)+24\cdot\left(99+1\right)}\)

\(A=\frac{12\cdot25+12\cdot175}{36\cdot100+24\cdot100}\)

\(A=\frac{12\cdot100}{60\cdot100}\)

\(A=\frac{1}{5}\)

Bấm L IKE ủng hộ nhá :)))

mik bt kết quả mà ko bt ách làm á bn
a) 2
b) 975

ko bt thì thôi muốn gây war hả bn

Sửa đề: \(180^0 và \(\sin x=-\frac45\)

\(180^0

=>sin x<0; cos x<0

Ta có: \(\sin^2x+cos^2x=1\)

=>\(cos^2x=1-\left(-\frac45\right)^2=1-\frac{16}{25}=\frac{9}{25}\)

mà cos x <0

nên \(cosx=-\frac35\)

\(cos\left(x-45^0\right)=cosx\cdot cos45^0+\sin x\cdot\sin45^0\)

\(=-\frac35\cdot\frac{1}{\sqrt2}+\frac{-4}{5}\cdot\frac{1}{\sqrt2}=\frac{-7}{5\sqrt2}=\frac{-7\sqrt2}{10}\)

a)

A=\(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5x-5}\)

\(\Leftrightarrow\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right)\div\dfrac{2x}{5\left(x-1\right)}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+1\\x=0-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

MTC: 5(x-1)(x+1)

\([\dfrac{5\left(x+1\right)\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}-\dfrac{5\left(x-1\right)\left(x-1\right)}{5\left(x-1\right)\left(x+1\right)}]\div\dfrac{2x\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow[5\left(x+1\right)\left(x+1\right)-5\left(x-1\right)\left(x-1\right)]\div2x\left(x+1\right)\)

\(\Leftrightarrow[5\left(x+1\right)^2-5\left(x-1\right)^2]\div2x^2+2x\)

\(\Leftrightarrow[5\left(x^2+2x+1\right)-5\left(x^2-2x+1\right)]\div2x^2+2x\)

\(\Leftrightarrow(5x^2+10x+5-5x^2+10x-5)\div2x^2+2x\)

\(\Leftrightarrow20x\div\left(2x^2+2x\right)\)

\(\Leftrightarrow10x+10\)

x+255=(-47)+45+255

         =-2+255
         =253

x+255=(-47)+45+255=(-47)+300=253

A=-25-24+24+-98

Trl:

\(A=m-24-x+24+x\)

\(A=m\)

\(\Rightarrow A=-25\)

Vậy ...

#HuyenAnh