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A=ba số hạng đầu
\(A=\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}=\frac{1}{x}-\frac{1}{x+6}\\ \)
B=3 số hạng tiếp theo
\(2B=\frac{1}{x+6}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}+\frac{1}{x+10}=\frac{1}{x+6}\)
\(A+B=\frac{1}{x}-\frac{1}{x+6}+\frac{1}{2\left(x+6\right)}=\frac{1}{x}-\frac{1}{2\left(x+6\right)}=\frac{12+x}{2x\left(x+6\right)}\)
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
i) (x - 1)(5x + 3) = (3x - 8)(x - 1)
<=> 5x2 + 3x - 5x - 3 = 3x2 - 3x - 8x + 8
<=> 5x2 - 2x - 3 = 3x2 - 11x + 8
<=> 5x2 - 2x - 3 - 3x2 + 11x - 8 = 0
<=> 2x2 + 9x - 11 = 0
<=> 2x2 + 11x - 2x - 11 = 0
<=> x(2x + 11) - (2x + 11) = 0
<=> (x - 1)(2x + 11) = 0
<=> x - 1 = 0 hoặc 2x + 11 = 0
<=> x = 0 hoặc x = -11/2
m) 2x(x - 1) = x2 - 1
<=> 2x2 - 2x = x2 - 1
<=> 2x2 - 2x - x2 + 1 = 0
<=> x2 - 2x + 1 = 0
<=> (x - 1)2 = 0
<=> x - 1 = 0
<=> x = 1
n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)
<=> 2x + 22 - 3x2 - 33x = 6x - 15x2 - 4 + 10x
<=> -31x + 22 - 3x2 = 16x - 15x2 - 4
<=> 31x - 22 + 3x2 + 16x - 15x2 - 4 = 0
<=> 47x - 18 - 12x2 = 0
<=> -12x2 + 47x - 26 = 0
<=> 12x2 - 47x + 26 = 0
<=> 12x2 - 8x - 39x + 26 = 0
<=> 4x(3x - 2) - 13(3x - 2) = 0
<=> (4x - 13)(3x - 2) = 0
<=> 4x - 13 = 0 hoặc 3x - 2 = 0
<=> x = 13/4 hoặc x = 2/3
a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)
\(3x=5\)
\(x=\frac{5}{3}\)
b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)
\(3x-8=2x-7\)
\(x=1\)
c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)
\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)
\(4x^2-3x-18=4x^2+3x\)
\(6x=-18\)
\(x=-3\)
d) Sai đề
e) ko bt
a)
\((x+2)(x+4)(x+6)(x+8)+16\)
\(=[(x+2)(x+8)][(x+4)(x+6)]+16\)
\(=(x^2+10x+16)(x^2+10x+24)+16\)
\(=a(a+8)+16\) (Đặt \(x^2+10x+16=a\) )
\(=a^2+2.4.a+4^2=(a+4)^2\)
\(=(x^2+10x+16+4)^2\)
\(=(x^2+10x+20)^2\)
b) \((x^2+x)(x^2+x+1)-6\)
\(=(x^2+x)^2+(x^2+x)-6\)
\(=(x^2+x)^2-2(x^2+x)+3(x^2+x)-6\)
\(=(x^2+x)(x^2+x-2)+3(x^2+x-2)\)
\(=(x^2+x-2)(x^2+x+3)\)
\(=(x^2-x+2x-2)(x^2+x+3)\)
\(=[x(x-1)+2(x-1)](x^2+x+3)\)
\(=(x-1)(x+2)(x^2+x+3)\)
c)
\((x^2-4x)^2-8(x^2-4x)+15\)
\(=(x^2-4x)^2-3(x^2-4x)-5(x^2-4x)+15\)
\(=(x^2-4x)(x^2-4x-3)-5(x^2-4x-3)\)
\(=(x^2-4x-3)(x^2-4x-5)\)
\(=(x^2-4x-3)(x^2+x-5x-5)\)
\(=(x^2-4x-3)[x(x+1)-5(x+1)]=(x^2-4x-3)(x+1)(x-5)\)
(-8+x^2)^5=1
<=>-8+x^2=1
<=>x^2=9
<=>x=3 hoặc -3
Vậy x=3 hoặc -3
Ta có: \(\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)\left(-8+x^2\right)=1\)
\(\Leftrightarrow\left(-8+x^2\right)^5=1\)
\(\Leftrightarrow x^2-8=\pm1\)
+ \(x^2-8=1\)\(\Leftrightarrow\)\(x^2=9\)\(\Leftrightarrow\)\(x=\pm3\)
+ \(x^2-8=-1\)\(\Leftrightarrow\)\(x^2=7\)\(\Leftrightarrow\)\(x=\pm\sqrt{7}\)
Vậy \(S=\left\{-3,-\sqrt{7},\sqrt{7},3\right\}\)