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Cho biểu thức: bn viết ko rõ lắm , bn xem đề mk viết lại có đg ko nhé , r mk lm cho
\(a=\dfrac{2x}{x+3}-\dfrac{x+1}{3-x}-\dfrac{3-11x}{x^2-9}\)
Bài 1:
\(A=\dfrac{1}{x-y}+\dfrac{1}{x+y}+\dfrac{2x}{x^2+y^2}+\dfrac{4x^3}{x^4+y^4}+\dfrac{8x^7}{x^8+y^8}\)
\(A=\dfrac{2x}{x^2-y^2}+\dfrac{2x}{x^2+y^2}+\dfrac{4x^3}{x^4+y^4}+\dfrac{8x^7}{x^8+y^8}\)
\(A=\dfrac{4x^3}{x^4-y^4}+\dfrac{4x^3}{x^4+y^4}+\dfrac{8x^7}{x^8+y^8}\)
\(A=\dfrac{8x^7}{x^8-y^8}+\dfrac{8x^7}{x^8+y^8}\)
\(A=\dfrac{16x^{15}}{x^{16}-y^{16}}\)
Câu 2:
a: \(n^2-2n+5⋮n-1\)
\(\Leftrightarrow n^2-n-n+1+4⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{2;0;3;-1;5;-3\right\}\)
b: \(4x^2-6x-16⋮x-3\)
\(\Leftrightarrow4x^2-12x+6x-18+2⋮x-3\)
\(\Leftrightarrow x-3\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{4;2;5;1\right\}\)
Câu 3:
a: \(\left(3x-8\right)\left(7x+10\right)-\left(2x-15\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(7x+10-2x+15\right)=0\)
\(\Leftrightarrow\left(3x-8\right)\left(5x+25\right)=0\)
=>x=8/3 hoặc x=-5
b: \(\dfrac{\left(x^4-2x^2-8\right)}{x-2}=0\)(ĐKXĐ: x<>2)
\(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2+2\right)=0\)
=>x+2=0
hay x=-2
a: \(C=\left(x+y\right)^2-2xy=6^2-2\cdot\left(-4\right)=36+8=44\)
\(D=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=288\)
b: \(A=x^2-6x+10=x^2-6x+9+1=\left(x-3\right)^2+1>0\)
\(B=x^2-2x+1+9y^2-6y+1+1=\left(x-1\right)^2+\left(3y-1\right)^2+1>0\)
c: \(A=x^2-4x+1=x^2-4x+4-3=\left(x-2\right)^2-3>=-3\)
Dấu = xảy ra khi x=2
\(B=4x^2+4x+1+10=\left(2x+1\right)^2+10>=10\)
Dấu = xảy ra khi x=-1/2
\(C=-\left(x^2+8x-5\right)\)
\(=-\left(x^2+8x+16-21\right)\)
\(=-\left(x+4\right)^2+21< =21\)
Dấu = xảy ra khi x=-4
\(D=-\left(x^2-5x\right)=-\left(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}\right)\)
\(=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}< =\dfrac{25}{4}\)
Dấu = xảy ra khi x=5/2
c)\(\left(xy^2-1\right)\left(x^2y+5\right)\)
\(=x^3y^3+5xy^2-x^2y-5\)
d)\(4\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x^2+1\right)\)
\(=4\left(x^2-\dfrac{1}{4}\right)\left(4x^2+1\right)\)
\(=4\left(4x^4+x^2-x-\dfrac{1}{4}\right)\)
\(=16x^4+4x^2-4x-1\)
M = -3(x – 4)(x – 2) + x(3x – 18) – 25
= -3( x 2 – 2x – 4x + 8) + x.3x + x.(-18) – 25
= -3 x 2 + 6x + 12x – 24 + 3 x 2 – 18x – 25
= (-3 x 2 + 3 x 2 ) + (6x + 12x – 18x) – 24 – 25
= -49
N = (x – 3)(x + 7) – (2x – 1)(x + 2) + x(x – 1)
= x.x + x.7 – 3.x – 3.7 – (2x.x + 2x.2 – x – 1.2) + x.x + x.(-1)
= x 2 + 7x – 3x – 21 – 2 x 2 – 4x + x + 2 + x 2 – x
= ( x 2 – 2 x 2 + x 2 ) + (7x – 3x – 4x + x – x) – 21 + 2
= -19
Vậy M = -49; N = -19 => M – N = -30
Đáp án cần chọn là: B
Ta có
M = 8(x – 1)( x 2 + x + 1) – (2x – 1)(4 x 2 + 2x + 1)
= 8( x 3 – 1) – ( 2 x 3 – 1)
= 8 x 3 – 8 – 8 x 3 + 1 = -7 nên M = -7
N = x(x + 2)(x – 2) – (x + 3)( x 2 – 3x + 9) – 4x
= x( x 2 – 4) – ( x 3 + 3 3 ) + 4x
= x 3 – 4x – x 3 – 27 + 4x = -27
=> N = -27
Vậy M = N + 20
Đáp án cần chọn là: D