(x-1)(x-3)(x-5)(x-7)=20
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20 x 1 + 20 x 2 + 20 x 3 + 20 x 4 + 20 x 5 + 20 x 6 = ?
1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x ... x 100 = ?
20x1+20x2+20x3+20x4+20x5+20x6= 20x(1+2+3+4+5+6)= 20 x 21 = 420
Còn câu 2 mik thấy khó quá cho mik xin lỗi nha
a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
=>1/x-1=3/4
=>x-1=4/3
=>x=7/3
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
1: Ta có: 7x+6(3-x)=27-20+73
\(\Leftrightarrow7x+18-6x=80\)
\(\Leftrightarrow x=80-18=62\)
Vậy: x=62
2: Ta có: \(6x-5\left(x-7\right)=\left(27-514\right)-486-73\)
\(\Leftrightarrow6x-5x+35=27-514-486-73\)
\(\Leftrightarrow x+35=-1046\)
\(\Leftrightarrow x=-1081\)
Vậy: x=-1081
Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14
2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84
3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1
5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5
6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3
\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)=20\\ \Leftrightarrow\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)\left[\left(x^2-8x+7\right)+8\right]-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)^2+8\left(x^2-8x+7\right)-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)^2-2\left(x^2-8x+7\right)+10\left(x^2-8x+7\right)-20=0\\ \Leftrightarrow\left(x^2-8x+7\right)\left(x^2-8x+7-2\right)+10\left(x^2-8x+7-2\right)=0\)
\(\Leftrightarrow\left(x^2-8x+7+10\right)\left(x^2-8x+7-2\right)=0\\ \Leftrightarrow\left(x^2-8x+17\right)\left(x^2-8x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-8x+16+1=0\\x^2-8x+16-11=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-4\right)^2+1=0\left(vô.lí\right)\\\left(x-4\right)^2-11=0\end{matrix}\right.\\ \Leftrightarrow\left(x-4-\sqrt{11}\right)\left(x-4+\sqrt{11}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{11}\\x=4-\sqrt{11}\end{matrix}\right.\)
\(\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-7\right)-20\)
\(=\left[\left(x-1\right)\left(x-7\right)\right]\left[\left(x-3\right)\left(x-5\right)\right]-20\)
\(=\left(x^2-8x+7\right)\left(x^2-8x+15\right)-20\)
Đặt \(x^2-8x+7=t\),ta có :
\(t\left(t+8\right)-20\)
\(=t^2+8t-20\)
\(=\left(t^2+8t+16\right)-16-20\)
\(=\left(t+4\right)^2-36\)
\(=\left(t+4\right)^2-6^2\)
\(=\left(t+4-6\right)\left(t+4+6\right)\)
\(=\left(t-2\right)\left(t+10\right)\)
\(=\left(x^2-8x+7-2\right)\left(x^2-8x+7+10\right)\)
\(=\left(x^2-8x+5\right)\left(x^2-8x+17\right)\)