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Tìm x biết :
\(\frac{1}{1×2×3×4}+\frac{1}{2×3×4×5}+\frac{1}{3×4×5×6}+...+\frac{1}{27×28×29×30}×x=-3\)
Ez thôi mà :)
B1: S = 1 + 2 + 3 + .. .+ n
=> S = ( n + 1 ) . n : 2 = aaa
=> S = ( n + 1 ) . n = 2aaa
Ta có: aaa = 111 . a = 37 . 3 . a
=> 2aaa = 37 . 6 . a
Mà ( n + 1 ) . n là 2 số tự nhiên liên tiếp => 6a = 36 => a = 6
=> ( n + 1 ) . n = 37 . 36
=> n = 36
B2: Đề sai thì phải -_- T sửa lại
(x + 2) + (4x + 4) + (7x + 6) + ... + (25x + 18) + (28x + 20) = 1560
<=> (x + 4x + 7x + ... + 25x + 28x) + (2 + 4 + 6 + ... + 18 + 20) = 1560
<=> 145x + 110 = 1560
<=> 145x = 1450
<=> x = 10
Câu 1 :
\(a,\left(3x+2\right)^2=9x^2+12x+4.\)
\(b,\left(6a^2-b\right)^2=36a^4-12a^2b-b^2\)
\(c,\left(4x-1\right)\left(4x+1\right)=16x^2-1\)
\(d,\left(1-x\right)\left(1+x\right)\left(1+x^2\right)=\left(1-x^2\right)\left(1+x^2\right)=1-x^4\)
\(e,\left(a^2+b^2\right)\left(a^2-b^2\right)=a^4-b^4\)
\(f,\left(x^3+y^2\right)\left(x^3-y^2\right)=x^6-y^4\)
Bài 2 :
\(a,A=9x^2+42x+49=9+42+49=100.\)
\(b,B=25x^2-2xy+\frac{1}{25}y^2=\left(5x^2\right)-2.5x.\frac{1}{5}y+\left(\frac{1}{5}y\right)^2\)
\(=\left(5x-\frac{1}{5}y\right)^2=\left(-1+1\right)^2=0\)
\(c,C=4x^2-28x+49=4x^2-14x-14x+49\)
\(=2x\left(x-7\right)-7\left(x-7\right)=\left(2x-7\right)\left(x-7\right)\)
\(=\left(8-7\right)\left(4-7\right)=-3\)
ta có y+4=(x-2)2=x2-4x+4 (1)
x+4=(y+2)2=y2-4y+4 (2)
Cộng (1)và (2), vế theo vế ta có :
x+y+8=x2-4x+4+y2-4y+4
\(\Rightarrow\) x2+y2=5x+5y
a) 3/4 + 1/4 : x = 2/5
1/4 : x = 2/5 - 3/4 = 8/20 - 15/20
1/4 : x = -7/20
x = 1/4 : -7/20 = 1/4 . -20/7
x = -5/7
b) 3/4 + 2/5 . x = 29/60
2/5 . x = 29/60 - 3/4 = 29/60 - 45/60
2/5 . x = -4/15
x= -4/15 : 2/ 5 = -4/5 . 5/2
x = -2/3
Ta có: \(f\left(x\right)=x^4+8x^3+23x^2+28x+12\)
=> \(f\left(x\right)=x^4+3x^3+5x^3+15x^2+8x^2+24x+4x+12\)
=> \(f\left(x\right)=x^3\left(x+3\right)+5x^2\left(x+3\right)+8x\left(x+3\right)+4\left(x+3\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left(x^3+5x^2+8x+4\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left(x^3+2x^2+3x^2+6x+2x+4\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left[x^2\left(x+2\right)+3x\left(x+2\right)+2\left(x+2\right)\right]\)
=> \(f\left(x\right)=\left(x+3\right)\left(x+2\right)\left(x^2+3x+2\right)\)
=> \(f\left(x\right)=\left(x+2\right)\left(x+3\right)\left(x^2+2x+x+2\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left(x+2\right)\left[x\left(x+2\right)+\left(x+2\right)\right]\)
=> \(f\left(x\right)=\left(x+3\right)\left(x+2\right)^2\left(x+1\right)\)
Cho \(A=1\cdot2+2\cdot3+...+28\cdot29+29\cdot30\)
\(3A=1\cdot2\cdot3-1\cdot2\cdot3+2\cdot3\cdot4-2\cdot3\cdot4+...+29\cdot30\cdot31\)
\(3A=29\cdot30\cdot31\)
\(A=\left(29\cdot30\cdot31\right):3=29\cdot10\cdot31=8990\)