(x^2-x)(x+2)(x+3)=18
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a) Ta có: \(\dfrac{x-2}{15}+\dfrac{x-3}{14}+\dfrac{x-4}{13}+\dfrac{x-5}{12}=4\)
\(\Leftrightarrow\dfrac{x-2}{15}-1+\dfrac{x-3}{14}-1+\dfrac{x-4}{13}-1+\dfrac{x-5}{12}-1=0\)
\(\Leftrightarrow\dfrac{x-17}{15}+\dfrac{x-17}{14}+\dfrac{x-17}{13}+\dfrac{x-17}{12}=0\)
\(\Leftrightarrow\left(x-17\right)\left(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}\right)=0\)
mà \(\dfrac{1}{15}+\dfrac{1}{14}+\dfrac{1}{13}+\dfrac{1}{12}>0\)
nên x-17=0
hay x=17
Vậy: x=17
b) Ta có: \(\dfrac{x+1}{19}+\dfrac{x+2}{18}+\dfrac{x+3}{17}+...+\dfrac{x+18}{2}+18=0\)
\(\Leftrightarrow\dfrac{x+1}{19}+1+\dfrac{x+2}{18}+1+\dfrac{x+3}{17}+1+...+\dfrac{x+18}{2}+1=0\)
\(\Leftrightarrow\dfrac{x+20}{19}+\dfrac{x+20}{18}+\dfrac{x+20}{17}+...+\dfrac{x+20}{2}=0\)
\(\Leftrightarrow\left(x+20\right)\left(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}\right)=0\)
mà \(\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+...+\dfrac{1}{2}>0\)
nên x+20=0
hay x=-20
Vậy: x=-20
![](https://rs.olm.vn/images/avt/0.png?1311)
(8 x 18 -5 x 18 -18 x3) x y + 2 x y =8 x 7 + 2
<=>144 x y -90 x y -54 x y + 2 x y = 58
<=>2 x y = 58
<=> y = 29
Vậy y = 29
Tìm y biết :
(8 x 18 - 5 x 18 - 18 x 3) x y + 2 x y = 8 x 7 + 2
[ 18 x ( 8 - 5 - 3 ) ] x y = 56 + 2
[ 18 x ( 8 - 5 - 3 ) ] x y = 58
( 18 x 0 ) x y = 58
0 x y + 2 x y = 58
2 x y = 58
y = 58 : 2
y = 29
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\(\frac{1.5.18+2.10.36+3.15.54}{1.3.9+2.6.18+3.9.27}=\frac{1.5.18.\left(1+2.2.2+3.3.3\right)}{1.3.9.\left(1+2.2.2+3.3.3\right)}\)
\(=\frac{1.5.18}{1.3.9}=\frac{10}{3}\)
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a) 32 : 4 : 2 = 8 : 2 = 4
32 : 4 : 2 = 32 : 2 = 16
b) 18 : 2 x 3 = 18 : 6 = 3
18 : 2 x 3 = 9 x 3 = 18
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c: =>x^3+9x^2+27x+27-x^2+6x-9-x^2+6x-9=6x+18
=>x^3+7x^2+39x+9-6x-18=0
=>x^3+7x^2+33x-9=0
=>\(x\simeq0.26\)
d: =>x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22
=>x^3-5x^2+2x-1=-5x^2-x-22
=>x^3+3x+21=0
=>\(x\simeq2.40\)
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Đề câu a có nhầm không nhỉ chứ lớp 8 chưa học phương trình vô tỉ ;-;
Không nhầm tag mình làm tiếp
a, Ta có : \(\left(x+2\right)^3-2\left(x-3\right)^2+18=0\)
\(\Leftrightarrow x^3+12x+6x^2+8-2x^2+12x-18+18=0\)
\(\Leftrightarrow x^3+4x^2+24x+8=0\)
b, Ta có : \(\left(x-5\right)^3-x\left(x-2\right)\left(x+2\right)+125=0\)
\(\Leftrightarrow x^3+75x-15x^2-125-x^3+4x+125=0\)
\(\Leftrightarrow-15x^2+79x=0\)
\(\Leftrightarrow x\left(-15x+79\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{71}{15}\end{matrix}\right.\)
Vậy ...
Câu a xem lại giúp ạ nghiệm rất xấu ;V
`b)(x-5)^3-x(x-2)(x+2)=-125`
`<=>x^3-15x^2+75x-125+125-x(x^2-4)=0`
`<=>x^3-15x^2+75x-x^3+4x^2=0`
`<=>75x-11x^2=0`
`<=>x(75-11x)=0`
`<=>` \(\left[ \begin{array}{l}x=0\\x=\dfrac{75}{11}\end{array} \right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-7.2\right)}\)
\(\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.-9}=\frac{2.9}{-9}=-2\)
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\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
![](https://rs.olm.vn/images/avt/0.png?1311)
(8 x 18 -5 x 18 -18 x3) x y + 2 x y =8 x 7 + 2
<=>144 x y -90 x y -54 x y + 2 x y = 58
<=>2 x y = 58
<=> y = 29
vậy y = 29
( mình ko chép lại đề bài đâu nha ,giải lun đó)
[18x(8-5-3)]xy+2xy=56+2
(18x0)xy+2xy=58
0xy+2xy=58
2xy=58
y=58:2=29
tick cho mình nha
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)\left(x+3\right)-18=0\)
\(\Leftrightarrow x\left(x+2\right)\left(x-1\right)\left(x+3\right)-18=0\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2+2x-3\right)-18=0\)
Đặt \(x^2+2x=t\)
\(\Rightarrow t^2-3t-18=0\Rightarrow\left[{}\begin{matrix}t=6\\t=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x=6\\x^2+2x=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x-6=0\\x^2+2x+3=0\end{matrix}\right.\) (casio)