x+3/-4=5/20
sai rồi
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a | b | c | a x (b + c) | a x b + a x c |
---|---|---|---|---|
4 | 5 | 2 | 4 x (5 + 2) = 28 | 4 x 5 + 4 x 2 = 28 |
3 | 4 | 5 | 3 x (4 + 5) = 27 | 3 x 4 + 3 x 5 = 27 |
6 | 2 | 3 | 6 x (2 + 3) = 30 | 6 x 2 + 6 x 3 = 30 |
a | b | c | a x ( b + c ) | a x b + a x c |
4 | 5 | 2 | 4 x ( 5 + 2 ) = 28 | 4 x 5 + 4 x 2 = 28 |
3 | 4 | 5 | 3 x ( 4 + 5 ) = 27 | 3 x 4 + 3 x 5 = 27 |
6 | 2 | 3 | 6 x ( 2 + 3 ) = 30 | 6 x 2 + 6 x 3 = 30 |
a | b | c | a x (b + c) | a x b + a x c |
---|---|---|---|---|
4 | 5 | 2 | 4 x (5 + 2) = 28 | 4 x 5 + 4 x 2 = 28 |
3 | 4 | 5 | 3 x (4 + 5) = 27 | 3 x 4 + 3 x 5 = 27 |
6 | 2 | 3 | 6 x (2 + 3) = 30 | 6 x 2 + 6 x 3 = 30 |
7/5+4/5 7/20-1/4 5/4x3/9 7/2 : 8
=7+4/5 =7/20-5/20 =5/4x1/3 =7/2x 1/8
=11/5 =7-5/20 =5/12 =7/16
= 1/10
Nhớ tick đấy
a) \(x-\dfrac{3}{5}=\dfrac{4}{-10}\)
\(x=\dfrac{4}{-10}+\dfrac{3}{5}\)
\(x=\dfrac{-4}{10}+\dfrac{6}{10}\)
\(x=\dfrac{1}{5}\)
b) \(\dfrac{3}{x}-2=\dfrac{4}{x}+4\)
\(\dfrac{3}{x}-2+2=\dfrac{4}{x}+4+2\)
\(\dfrac{3}{x}=\dfrac{4}{x}+4\)
\(\dfrac{3}{x}=\dfrac{4x+4}{x}\)
\(3x=\left(4x+4\right)x\)
\(3x=5x\cdot x+4x\)
\(3x=x\left(5x+4\right)\)
\(3=5x+4\)
\(5x=-1\)
\(x=\dfrac{-1}{5}\)
a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)
\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)
\(=-5x^3-38x^2-76x-46\)
b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)
\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)
\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)
\(=x^3-20x^2+58x+69\)
c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)
\(=-18x^3-46x^2-8x+16\).
a) Ta có: \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)
\(=x^3+2x^2-9x-18-\left(x^3-3x-x^2+3\right)-5x^3-40x^2-80x-x^2+10x-25\)
\(=-4x^3-39x^2-79x-43-x^3+3x+x^2-3\)
\(=-5x^3-38x^2-76x-46\)
b) Ta có: \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)
\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2x^2+20x+50-x^2+2x-1\)
\(=2x^3-16x^2+32x-x^3+4x-5x^2+20+x^2+22x+49\)
\(=x^3-20x^2+56x+49\)
c) Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3+48x-36x-2x^3+18x+x^2-9\)
\(=-18x^3+2x^2+40x+16\)
Ta có: x + 3/-4 = 5/20.
=> x - 3/4 = 1/4.
x = 1/4 + 3/4.
x = 1
là \(\frac{x+3}{-4}\) hay là \(x+\frac{3}{-4}\) vậy nguyễn đức tín