Cho hinh thang ABCD ( BC//AD). Biet BC+AD=AB.CM:Cac tia phan giac cua goc A va B cat nhau tai trung diem canh CD
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\(25x^2-16=0=>\left(5x\right)^2-4^2=0=>\left(5x-4\right)\left(5x+4\right)=0\)
\(=>\orbr{\begin{cases}5x-4=0\\5x+4=0\end{cases}=>\orbr{\begin{cases}5x=4=>x=\frac{4}{5}\\5x=-4=>x=-\frac{4}{5}\end{cases}}}\)
Từ \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0=>\frac{ayz}{xyz}+\frac{bxz}{xyz}+\frac{cxy}{xyz}=0=>\frac{ayz+bxz+cxy}{xyz}=0=>ayz+bxz+cxy=0\)
Từ \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1=>\left(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\right)^2=1^2\)
\(=>\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}+2\left(\frac{xy}{ab}+\frac{yz}{bc}+\frac{xz}{ac}\right)=1\)
\(=>\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1-2\left(\frac{xyc}{abc}+\frac{yza}{abc}+\frac{xzb}{abc}\right)=1-2.0=1\)
Vậy M=1
a) \(A=\left(x^3+x^2\right)-\left(x+1\right)=x\left(x+1\right)-\left(x+1\right)=\left(x-1\right)\left(x+1\right)\)
b) \(B=\left(x^3-3x^2\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
đề sai nà : sử lại x.(x+1)(x+2)(x+3)+1
Ta có
x(x+3)=x2+3x
(x+1)(x+2)=x2+x+2x+2=x2+3x+2
=> x.(x+1)(x+2)(x+3)+1=(x2+3x)(x2+3x+2)+1
Đặt y=x2+3x=>y+2=x2+3x+2
=> x.(x+1)(x+2)(x+3)+1=y(y+2)+1=y2+2y+1=(y+1)2=(x2+3x+1)2
T i c k cho mình nha cảm ơn nhìu
\(x^2-4x+3=x^2-3x-x+3=x\left(x-3\right)-\left(x-3\right)=\left(x-1\right)\left(x-3\right)\)
\(x^2+5x+4=x^2+4x+x+4=x\left(x+4\right)+\left(x+4\right)=\left(x+1\right)\left(x+4\right)\)
\(x^2-x-6=x^2-3x+2x-6=x\left(x-3\right)+2\left(x-3\right)=\left(x+2\right)\left(x-3\right)\)
\(x^4+4=\left(x^2\right)^2+2.x^2.2+2^2-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2\right)^2-\left(2x^2\right)=\left(x^2+2+2x\right)\left(x^2-2-2x\right)\)