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1
Với \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\)
\(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\left(\dfrac{x^2+2x+1}{4x^4-4x^2+1}\right)\\ =\left(\dfrac{\left(x-1\right)\left(x+1\right)}{\left(2-x\right)\left(x+1\right)}+\dfrac{x^2}{\left(x+1\right)\left(2-x\right)}\right)\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{x^2-1+x^2}{\left(x+1\right)\left(2-x\right)}\left(\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\right)\\ =\dfrac{\left(2x^2-1\right)\left(x+1\right)^2}{\left(x+1\right)\left(2-x\right)\left(2x^2-1\right)^2}\\ =\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}\)
2
Để M = 0 thì \(\dfrac{x+1}{\left(2-x\right)\left(2x^2-1\right)}=0\Rightarrow x+1=0\Rightarrow x=-1\) (loại)
Vậy không có giá trị x thỏa mãn M = 0
1) \(M=\left(\dfrac{x-1}{2-x}-\dfrac{x^2}{x^2-x-2}\right)\cdot\dfrac{x^2+2x+1}{4x^4-4x^2+1}\) (ĐK: \(\left\{{}\begin{matrix}x\ne2\\x\ne-1\\x\ne\sqrt{\dfrac{1}{2}}\end{matrix}\right.\))
\(M=\left(\dfrac{-\left(x-1\right)}{x-2}-\dfrac{x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-\left(x-1\right)\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-\left(x^2-1\right)-x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\left(\dfrac{-x^2+1-x^2}{\left(x-2\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\dfrac{-2x^2+1}{\left(x-2\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{\left(2x^2-1\right)^2}\)
\(M=\dfrac{-\left(2x^2-1\right)\left(x+1\right)^2}{\left(x-2\right)\left(x+1\right)\left(2x^2-1\right)^2}\)
\(M=\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}\)
2) Ta có: \(M=0\)
\(\Rightarrow\dfrac{-\left(x+1\right)}{\left(x-2\right)\left(2x^2-1\right)}=0\)
\(\Leftrightarrow-\left(x+1\right)=0\)
\(\Leftrightarrow-x=1\)
\(\Leftrightarrow x=-1\left(ktm\right)\)
b: \(N=a^3-3a^2-a\left(3-a\right)\)
\(=a^2\left(a-3\right)+a\left(a-3\right)\)
\(=a\left(a-3\right)\left(a+1\right)\)
a,theo giả thiết E lần lượt là hình chiếu của H lên AB,
H là chân đường vuông góc kẻ từ B xuống AC
\(=>\)\(\angle\left(BEH\right)=\angle\left(BHA\right)=90^o\)
có \(\angle\left(B\right)chung\)\(=>\Delta BEH\sim\Delta BHA\left(g.g\right)\left(dpcm\right)\)
b, ta có E,F là hình chiếu của H trên AB,BC
\(=>HE\perp AB,HF\perp BC\)
mà \(BH\perp AC\left(gt\right)=>\)\(\Delta BHA\) vuông tại H có HE là đường cao
và \(\Delta BHC\) vuông tại H có HF là đường cao
theo hệ thức lượng
\(=>BH^2=BE.BA=BF.BC\left(dpcm\right)\)
`a)P(x)+Q(x)=x^5-2x^2+1`
`=>Q(x)=x^5-2x^2+1-P(x)`
`=>Q(x)=x^5-2x^2+1-x^4+3x^2-1/2+x`
`=>Q(x)=x^5-x^4+x^2+x+1/2`
______________________________________________
`b)P(x)-R(x)=x^3`
`=>R(x)=P(x)-x^3`
`=>R(x)=x^4-3x^2+1/2-x-x^3`
`=>R(x)=x^4-x^3-3x^2-x+1/2`
Ta có:
\(P\left(x\right)+Q\left(x\right)=x^5-2x^2+1\)
\(\Rightarrow Q\left(x\right)=P\left(x\right)-\left(x^5-2x^2+1\right)\)
\(=x^4-3x^2+\dfrac{1}{2}-x-x^5+2x^2-1\)
\(=-x^5+x^4-x^2-x-\dfrac{1}{2}\)
Vậy \(Q\left(x\right)=-5^2+x^4-x^2-x-\dfrac{1}{2}\)
\(4x^2-5x+1=4x^2-4x-x+1=4x\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(4x-1\right)\)
\(\left(y-5\right)\left(y+8\right)-\left(y+4\right)\left(y-1\right)\)
\(=y^2+3y-40-y^2-3y+4=-36\)
Bạn phải cho x-y mới tìm đc
\(x+y=3\Rightarrow\left(x+y\right)^2=9\Rightarrow x^2+y^2+2xy=9\)
\(\Rightarrow5+2xy=9\Rightarrow xy=2\)
\(\Rightarrow\left(x-y\right)^2=x^2+y^2-2xy=5-2.2=1\)
\(\Rightarrow x-y=\pm1\)
\(A=\left(x-y\right)\left(x^2+y^2+xy\right)=\pm1.\left(5+2\right)=\pm7\)