Bài 1 : Nếu :
\(\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)=1\) \(=1\)
Tìm x
2. Tìm x biết:
a)2(x+2)(x+4)\dfrac{2}{\left(x+2\right)\left(x+4\right)} + 4(x+4)(x+8)\dfrac{4}{\left(x+4\right)\left(x+8\right)} + 6(x+8)(x+14)\dfrac{6}{\left(x+8\right)\left(x+14\right)} = x(x+2)(x+14)\dfrac{x}{\left(x+2\right)\left(x+14\right)}
b)x2023\dfrac{x}{2023} + x+12022\dfrac{x+1}{2022} x+22021\dfrac{x+2}{2021} +...+ x+20221\dfrac{x+2022}{1} + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}\)+\(\dfrac{2}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
Sửa đề:
\(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
ĐKXĐ: \(x\notin\left\{1;3;8;20\right\}\)
PT=>\(-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=-\dfrac{3}{4}\)
=>\(-\dfrac{1}{x-4}=-\dfrac{3}{4}\)
=>\(x-1=\dfrac{4}{3}\)
=>\(x=\dfrac{4}{3}+1=\dfrac{7}{3}\)(nhận)
Tìm x biết:
a) \(\left|x+2\dfrac{1}{2}\right|=\left|3x+1\right|\)
b) \(\left|2x-6\right|+\left|x+3\right|=8\)
c) \(2.\left|x+2\right|+\left|4-x\right|=11\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
\(b,\Rightarrow\left[{}\begin{matrix}6-2x-x-3=8\left(x\le-3\right)\\6-2x+x+3=8\left(-3\le x\le3\right)\\2x-6+x+3=8\left(x>3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\left(ktm\right)\\x=1\left(tm\right)\\x=\dfrac{11}{3}\left(tm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{3}\end{matrix}\right.\)
\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)\(\sqrt{ }\)\(\left|2x+3\right|+\left|2x-1\right|\)=\(\dfrac{8}{3.\left(x+1\right)^2+2}\)
Tìm \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Bài 1 : Phân tích đa thức thành nhân tử
\(a,5x\left(x-2y\right)+2\left(2y-x\right)^2\)
\(b,7x\left(y-4\right)^2-\left(4-x\right)^3\)
\(c,\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
Bài 2 :
a, \(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
b, \(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{\left(x-20\right)}=\frac{-3}{4}\)
\(\frac{2}{\left(x-1\right).\left(x-3\right)}+\frac{5}{\left(x-3\right).\left(x-8\right)}+\frac{12}{\left(x-8\right).\left(x-20\right)}-\frac{1}{x-20}=\frac{-3}{4}\)
Tìm x
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
ngu như con bò tót, ko biết 1+1=2.
\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\frac{1}{x-1}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-8}+\frac{1}{x-8}-\frac{1}{x-20}-\frac{1}{x-20}=-\frac{3}{4}\)
\(\frac{1}{x-1}-\frac{2}{x-20}=-\frac{3}{4}\)
\(\frac{x-20-2x+2}{x^2-21x+20}=\frac{-3}{4}\)
\(-4x-72=-3x^2+63x-60\)
\(-3x^2+63x-60+4x+72=0\)
\(-3x^2+67x+12=0\)
Ko có x t/m