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\(x^4-9x^3+x^2-9x\\ =x^3\left(x-9\right)+x\left(x-9\right)\\ =\left(x^3+x\right)\left(x-9\right)=x\left(x^2+1\right)\left(x-9\right)\\ ---\\ x^2-9-x^2\left(x^2-9\right)\\ =\left(x^2-9\right)-x^2\left(x^2-9\right)\\ =\left(x^2-9\right)\left(1-x^2\right)=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\\ ---\\ xz+yz-5x-5y=z\left(x+y\right)-5\left(x+y\right)\\ =\left(z-5\right)\left(x+y\right)\\ ---\\ x^3-2x+y^3-2y\\ =\left(x^3+y^3\right)-2\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2-2\right)\)
\(45+x^3-5x^2-9x\\ =x^2\left(x-5\right)-9\left(x-5\right)\\ =\left(x^2-9\right)\left(x-5\right)=\left(x-3\right)\left(x+3\right)\left(x-5\right)\\ ---\\ x^2-y^2-2x-2y\\ =\left(x^2-y^2\right)-\left(2x+2y\right)\\ =\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\\ =\left(x+y\right)\left(x-y-2\right)\\ ---\\ x^3+7x^2-4x-28\\ =x^2\left(x+7\right)-4\left(x+7\right)\\ =\left(x+7\right)\left(x^2-4\right)=\left(x+7\right)\left(x-2\right)\left(x+2\right)\\ ---\\ 8x^3-y^3-4x+2y=\left(8x^3-y^3\right)-2\left(2x-y\right)\\ =\left(2x-y\right)\left(4x^2+2xy+y^2\right)-2\left(2x-y\right)\\ =\left(2x-y\right)\left(4x^2+2xy+y^2-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 5 hình 1: (tự vẽ hình nhé bạn)
a) Xét ΔABD và ΔACB ta có:
\(\widehat{BAD}\)= \(\widehat{BAC}\) (góc chung)
\(\widehat{ABD}\)= \(\widehat{ACB}\) (gt)
=> ΔABD ~ ΔACB (g-g)
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CB}\) = \(\dfrac{AD}{AB}\) (tsđd)
b) Ta có: \(\dfrac{AB}{AC}\) = \(\dfrac{AD}{AB}\) (cm a)
=> \(AB^2\) = AD.AC
=> \(2^2\) = AD.4
=> AD = 1 (cm)
Ta có: AC = AD + DC (D thuộc AC)
=> 4 = 1 + DC
=> DC = 3 (cm)
c) Xét ΔABH và ΔADE ta có:
\(\widehat{AHB}\) = \(\widehat{AED}\) (=\(90^0\))
\(\widehat{ADB}\) = \(\widehat{ABH}\) (ΔABD ~ ΔACB)
=> ΔABH ~ ΔADE
=> \(\dfrac{AB}{AD}\) = \(\dfrac{AH}{AE}\) = \(\dfrac{BH}{DE}\) (tsdd)
Ta có: \(\dfrac{S_{ABH}}{S_{ADE}}\) = \(\left(\dfrac{AB}{AD}\right)^2\)= \(\left(\dfrac{2}{1}\right)^2\)= 4
=> đpcm
Tiếp bài 5 hình 2 (tự vẽ hình)
a) Xét ΔABC vuông tại A ta có:
\(BC^2\) = \(AB^2\) + \(AC^2\)
\(BC^2\) = \(21^2\) + \(28^2\)
BC = 35 (cm)
b) Xét ΔABC và ΔHBA ta có:
\(\widehat{BAC}\) = \(\widehat{AHB}\) ( =\(90^0\))
\(\widehat{ABC}\) = \(\widehat{ABH}\) (góc chung)
=> ΔABC ~ ΔHBA (g-g)
=> \(\dfrac{AB}{BH}\) = \(\dfrac{BC}{AB}\) (tsdd)
=> \(AB^2\) = BH.BC
=> \(21^2\) = 35.BH
=> BH = 12,6 (cm)
c) Xét ΔABC ta có:
BD là đường p/g (gt)
=> \(\dfrac{AD}{DC}\) = \(\dfrac{AB}{BC}\) (t/c đường p/g)
Xét ΔABH ta có:
BE là đường p/g (gt)
=> \(\dfrac{HE}{AE}\) = \(\dfrac{BH}{AB}\) (t/c đường p/g)
Mà: \(\dfrac{AB}{BC}\) = \(\dfrac{BH}{AB}\) (cm b)
=> đpcm
d) Ta có: \(\left\{{}\begin{matrix}\widehat{HBE}+\widehat{BEH}=90^0\\\widehat{ABD}+\widehat{ADB=90^0}\\\widehat{HBE}=\widehat{ABD}\end{matrix}\right.\)
=> \(\widehat{BEH}=\widehat{ADB}\)
Mà \(\widehat{BEH}=\widehat{AED}\) (2 góc dd)
Nên \(\widehat{ADB}=\widehat{AED}\)
=> đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
Số đo góc ngoài tại đỉnh D là:
\(180^0-360^0+70^0+90^0+120^0=100^0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(2x+x^2=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
b) Ta có: \(\left(2x+1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-4\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(5x+10y=5\left(x+2y\right)\)
b) \(3x^2y+9xy^2z=3xy\left(x+3yz\right)\)
g) \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
h) \(x^2+9x+8=\left(x+8\right)\left(x+1\right)\)
l) \(x^2-10x+9=\left(x-1\right)\left(x-9\right)\)
k) \(x^2+x-12=\left(x+4\right)\left(x-3\right)\)
l) \(3x^2+8x+4=\left(3x+2\right)\left(x+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(2x+3\right).5x=10x^2.15x\)
\(b,1011^2-1010^2=\left(1011-1010\right)\left(1011+1010\right)=2021\)
\(c,x^2+3x=x\left(x+3\right)\)
\(c,x^2+2xy-x-2y=\left(x^2-x\right)+\left(2xy-2y\right)=x\left(x-1\right)+2y\left(x-1\right)=\left(x-1\right)\left(x+2y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Ta có: \(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-60x^2+15x=-100\)
\(\Leftrightarrow50x=-100\)
hay x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x^3+2\)
\(=x^3-1-x^3+2\)
=1
b: ta có: \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)-8x^3-5\)
\(=8x^3-y^3-8x^3-5\)
\(=-y^3-5\)
c: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x^3+2\)
\(=x^3+1-x^3+2\)
=3
d: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-8x^3-5\)
\(=8x^3+y^3-8x^3-5\)
\(=y^3-5\)
a) \(\left(x-1\right)\left(x^2+x+1\right)-x^3+2=x^3-1-x^3+2=1\)
b) \(\left(2x-y\right)\left(4x^2+2xy+y^2\right)-8x^3-5=8x^3-y^3-8x^3-5=-y^3-5\)
c) \(\left(x+1\right)\left(x^2-x+1\right)-x^3+2=x^2+1-x^3+2=3\)
d) \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-8x^3-5=8x^3+y^3-8x^3-5=y^3-5\)
e) \(\left(3x+2\right)\left(9x^2-6x+4\right)-27x^3-7=27x^3+8-27x^3-7=1\)
f) \(\left(3x-2\right)\left(9x^2+6x+4\right)-27x^3-7=27x^3-8-27x^3-7=-15\)
a) \(=x^3-8\)
b) \(=\left(2x\right)^3+1=8x^3+1\)
c) \(=1-\left(\dfrac{x}{2}\right)^3=1-\dfrac{x^3}{8}\)
d) \(=y^3-\left(\dfrac{x}{y}\right)^3\)
a) \(M=x^3+27\)
b) \(N=1-\left(3x\right)^3=1-27x^3\)
c) \(P=x^3-\left(\dfrac{1}{2}\right)^3=x^3-\dfrac{1}{8}\)
d) \(Q=\left(2x\right)^3+\left(3y\right)^3=8x^3+27y^3\)
a: (x-2)(x^2+2x+4)=x^3-2^3=x^3-8
b; (2x+1)(4x^2-2x+1)
=(2x)^3+1^3
=8x^3+1
c: (1-x/2)(1+x/2+x^2/4)
=1-(x/2)^3
=1-1/8x^3
d: (y-x/y)(y^2+x+x^2/y^2)
=y^3-(x/y)^3
=y^3-x^3/y^3
a: (x+3)(x^2-3x+9)
=x^3+3^3
=x^3+27
b: (1-3x)(1+3x+9x^2)
=1^3-(3x)^3
=1-27x^3
c: (x-1/2)(x^2+1/2x+1/4)
=x^3-(1/2)^3
=x^3-1/8
d: (2x+3y)(4x^2-6xy+9y^2)
=(2x)^3+(3y)^3
=8x^3+27y^3