1/2 + 1/6 + 1/12 +.........1/x(x+1)
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Chứng minh phương trình x^2-4x+7=0 vô nghiệmGiải các phương trình sau: a)x+1/x^2+7x+12 +1/x^2+3x+2 +1/x^2+5x+6 +1/x^2+7x+12 +1/x^2+9x+20 +1/x^2+11x+30=8x-7/x(x-6)b) x+1/999 +x+2/998 = x+3/997 +x+4/996c) 6/x^2+4x+3 +6/x^2+8x+15 +6/x^2+12...
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x2-4x+7 = 0 ⇔ x2 -4x + 4 + 3 = 0
⇔ (x-2)2+3=0 ⇔ (x-2)2=-3 (vô lí)
Vậy pt vô nghiệm
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*Chứng minh phương trình \(x^2-4x+7=0\) vô nghiệm
Ta có: \(x^2-4x+7=0\)
\(\Leftrightarrow x^2-4x+4+3=0\)
\(\Leftrightarrow\left(x-2\right)^2+3=0\)
mà \(\left(x-2\right)^2+3\ge3>0\forall x\)
nên \(x\in\varnothing\)(đpcm)
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2*x+1/2*x + 6*x+1/6*x +12*x+1/12*x +20*x+1/20*x= 5.tìm x
2 x X + 1/2 + 4 x X + 1/6 + 6 x X + 1/6 = 12 11/12 . tìm x
\(2x+\dfrac{1}{2}+4x+\dfrac{1}{6}+6x+\dfrac{1}{6}=12\dfrac{11}{12}\Leftrightarrow\)\(2x+\dfrac{1}{2}+4x+\dfrac{1}{6}+6x+\dfrac{1}{6}=\dfrac{155}{12}\)\(\Leftrightarrow24x+6+48x+2+72x+2=155\)\(\Leftrightarrow24x+48x+72x=155-6-2-2\)\(\Leftrightarrow144x=145\)\(\Leftrightarrow x=\dfrac{145}{144}\)
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2*x+1/2*x+6*x+1/6*x+12*x+1/12*x+20*x+1/20*x=5.Tim x
a.A1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700
Đọc tiếp
a.A=1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B=1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C=1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700
1/2 + 1/6 + 1/12 + ... + 1/ X x (x+1) = 1/2
bài tìm x
\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\times\left(x+1\right)}=\dfrac{1}{2}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{x\times\left(x+1\right)}=\dfrac{1}{2}\)
\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{1}{2}\)
\(1-\dfrac{1}{x+1}=\dfrac{1}{2}\)
\(\dfrac{1}{x+1}=1-\dfrac{1}{2}=\dfrac{2}{2}-\dfrac{1}{2}\)
\(\dfrac{1}{x+1}=\dfrac{1}{2}\)
\(=>x+1=2\)
\(x=2-1\)
\(=>x=1\)
\(#WendyDang\)
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a) \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)
b) \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
a/ \(\dfrac{1}{x^2+x}+\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}\)
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+4}\)
Vậy..
b/ \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}\)
\(=\dfrac{1}{x+1}-\dfrac{1}{x+5}\)
Vậy..
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Tìm x, biết 12 chia hết cho x và a -2A.
x
∈
−
1
.
B.
x
∈
−
3
;
−
4
;
−
6
;
−
12
.
C.
x...
Đọc tiếp
Tìm x, biết 12 chia hết cho x và a < -2
A. x ∈ − 1 .
B. x ∈ − 3 ; − 4 ; − 6 ; − 12 .
C. x ∈ − 2 ; − 1 .
D. x ∈ − 2 ; − 1 ; 1 ; 2 ; 3 ; 46 ; 12 .
Đáp án là B vì 12: -3 = -4; 12: -4 = -3; 12: -6 = -2;12: -12 = -1 và đáp ứng điều kiện a< -2
a.A=1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B=1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C=1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700
a.A=1 1/2 x 1 1/3 x 1 1/4 x 1 1/5 x...x1 1/2020 x 1 1/2021
b.B=1 1/2 + 1 1/6 + 1 1/12 +1 1/20+ ... + 1 1/380 + 1 1/420
c.C=1 6/8 x 1 6/18 x 1 6/30 x 1 6/44 x ... x 1 6/10700