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a)
\(P=\dfrac{x^{10}-x^8+x^6-x^4+x^2-1}{x^4-1}\)
\(=\dfrac{x^8\left(x^2-1\right)+x^4\left(x^2-1\right)+\left(x^2-1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\dfrac{\left(x^2-1\right)\left(x^8+x^4+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}\)
\(=\dfrac{x^8+x^4+1}{x^2+1}\)
b)
\(Q=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{\left(x^{45}+x^{35}+...+x^5\right)+\left(x^{40}+x^{30}+...+1\right)}\)
\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^5\left(x^{40}+x^{30}+...+1\right)+\left(x^{40}+x^{30}+...+1\right)}\)
\(=\dfrac{x^{40}+x^{30}+x^{20}+x^{10}+1}{\left(x^{40}+x^{30}+...+1\right)\left(x^5+1\right)}\)
\(=\dfrac{1}{\left(x^5+1\right)}\)
cái câu b dòng cuối mẫu số đóng mở ngoặc chi cho mệt ei =.=
a) 2x (x+3) - 2 (x+2)2
= 2x2+6 - 2 ( x2+4x+4)
= 2x2+6 - 2x2 - 8x - 8
= -8x - 2
b) 2(x+2)(x-2) -(2x-3)(x-1)
= 2(x2-4)-2x2+2x+ 3x - 3
= 2x2 - 8 - 2x2+2x+ 3x - 3
= 5x - 11
c) (x+1)2 -(2x2+6)-(x+1)(x-1)
= x2 +2x+1 - 2x2 -6 - x2+1
= -2x2 + 2x
a, đk : x khác 5;-6
\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm)
b, đk : x khác 1;3
\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)
pt vô nghiệm
a, đk : x khác 5;-6
x2+12x+36+x2−10x+25=2x2+23x+61x2+12x+36+x2−10x+25=2x2+23x+61
⇔2x+61=23x+61⇔21x=0⇔x=0⇔2x+61=23x+61⇔21x=0⇔x=0(tm)
b, đk : x khác 1;3
x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)
pt vô nghiệm
Đặt biểu thức là A, ta có:
\(A=\frac{x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(\Rightarrow A.x^5=\frac{x^{45}+x^{35}+x^{25}+x^{15}+x^5}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(\Rightarrow A.x^5+A=\frac{x^{45}+x^{40}+x^{35}+x^{25}+x^{15}+x^5+x^{40}+x^{30}+x^{20}+x^{10}+1}{x^{45}+x^{40}+x^{35}+...+x^{10}+x^5+1}\)
\(\Rightarrow A.x^5+1=1\)
\(\Rightarrow A=\frac{1}{x^5+1}\)
\(1,=\left(x+3\right)\left(x-2\right):\left(x+3\right)=x-2\\ 2,=\left(x-5\right)\left(x+6\right):\left(x+6\right)=x-5\\ 3,=\left[3x\left(2x-1\right)-5\right]:\left(2x-1\right)=3x.dư.\left(-5\right)\)
1)\(\left(x+x^2-6\right):\left(x+3\right)=\left[x\left(x+3\right)-2\left(x+3\right)\right]:\left(x+3\right)=\left[\left(x+3\right)\left(x-2\right)\right]:\left(x+3\right)=x-2\)
2) \(\left(x+x^2-30\right):\left(x+6\right)=\left[x\left(x+6\right)-5\left(x+6\right)\right]:\left(x+6\right)=\left[\left(x+6\right)\left(x-5\right)\right]:\left(x+6\right)=x-5\)
3) \(\left(5-3x+6x^2\right):\left(2x-1\right)=\left[3x\left(2x-1\right)+5\right]:\left(2x-1\right)=3x+\dfrac{5}{2x-1}\)
mik chẳng hỉu gì cả
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lớp 8 ???
\(\dfrac{x}{30}+\dfrac{1}{2}+\dfrac{x}{40}=5+\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{x}{30}+\dfrac{x}{40}=\dfrac{14}{3}\)
\(\Leftrightarrow\dfrac{4x+3x}{120}=\dfrac{560}{120}\)
\(\Leftrightarrow7x=560\)
\(\Leftrightarrow x=80\)