2/3-1 và 4/5x = -3/5
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a) Cậu xem lại đề đi
b) \(3x.\left(x-2\right)-5x.\left(1-x\right)-8.\left(x^2-3\right)=4\)\(\Leftrightarrow3x^2-6x-5x+5x^2-8x^2+24-4=0\Leftrightarrow-11x+20=0\Leftrightarrow-11x=-20\Leftrightarrow x=\frac{20}{11}\)
c) \(2x^2+3.\left(x-1\right)\left(x+1\right)=5x\left(x+1\right)\Leftrightarrow2x^2+3\left(x^2-1\right)-5x\left(x+1\right)=0\)
\(\Leftrightarrow2x^2+3x^2-3-5x^2-5x=0\Leftrightarrow-5x=3\Leftrightarrow x=-\frac{3}{5}\)
Trần Anh: Cảm ơn bạn nhiều nhé :)) Phần a đúng là có sai đề pạn ạ mik làm hoài mà cux ko ra hì hì !!~~ Dù sao mik cux cảm ơn pạn nhiều nhiều nhé :3
a)\(P=\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)\)
\(P=5x-1+5x+4+\left(2-10x\right)\left(4+5x\right)\)
\(P=10x+3+8+10x-40x-50x^2\)
\(P=-20x+11-50x^2\)
b)\(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(Q=x^3-3x^2y+3xy^2-y^3+y^3+3y^2x+3yx^2+x^3+y^3-3y^2x+3yx^2-x^3-3x^2y+3xy^2\)
\(Q=x^3+y^3\)
a) P = (5x - 1) + [(-2).(-1).(1-5x).(4+5x)] + (5x+4)^2
= (5x - 1).[(-1)-2.(-1).(4+5x)] + (25x^2+40x+16)
= [(5x - 1).(10x + 7)] + (25x^2+40x+16)
= 50x^2 +35x - 10x - 7 + 25x^2+40x+16
= 75x^2 + 65x -9
b) Q = (x-y)^3 + (x+y)^3 + (y-x)^3 - [3xy(x+y)]
Xét (x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3 = x^3 + 3x^2y + 3xy^2 + y^3 = y^3 - 3xy^2 + 3x^2y - x^3
Do đó Q = (2x^3 + 6xy^2) + (y^3 - 3xy^2 + 3x^2y - x^3) - [3xy(x+y)]
= (x^3 + 3xy^2 + 3x^2y + y^3) - 3x^2y - 3xy^2
= x^3 + y^3
a, P=(5x-1)+2(1-5x)(4+5x)+(5x+4)
=5x-1+2.(4-15x-5x2)+5x+4
=5x-1+8-30x-10x2+5x+4
=-10x2-20x+11
b,Q=(x-y)^3 +(y+x)^3+ (y-x)^3 -3xy(x+y)
=x3-3x2y+3xy2-y3+y3+3x2y+3xy2+x3+y3-3y2x+3yx2-x3-3x2y-3xy2
=x3+y3
a)
P=(5x−1)+2(1−5x)(4+5x)+(5x+4)2
P=(5x−1)2+2(1−5x)(4+5x)+(5x+4)2−(5x−1)2
P=(5x−1+5x+4)2−(5x−1)2
P=(5x−1+5x+4−5x+1)(5x−1+5x+4+5x−1
b)
Q= (x-y)^3 + (x+y)^3 + (y-x)^3 - [3xy(x+y)]
(x-y)^3 = x^3 - 3x^2y + 3xy^2 - y^3
(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3
(y-x)^3 = y^3 - 3xy^2 + 3x^2y - x^3
Q= (2x^3 + 6xy^2) + (y^3 - 3xy^2 + 3x^2y - x^3) - [3xy(x+y)]
= (x^3 + 3xy^2 + 3x^2y + y^3) - 3x^2y - 3xy^2
Q= x^3 + y^3
`@` `\text {Ans}`
`\downarrow`
`1)`
\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(2x=\dfrac{7}{6}\)
`\Rightarrow`\(x=\dfrac{7}{6}\div2\)
`\Rightarrow`\(x=\dfrac{7}{12}\)
Vậy, `x = 7/12`
`2)`
\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)
`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)
`\Rightarrow`\(x=\dfrac{40}{21}\)
Vậy, `x = 40/21`
`3)`
\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)
`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)
`\Rightarrow`\(x=\dfrac{16}{21}\)
Vậy, `x = 16/21`
`4)`
\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{1}{12}\)
`\Rightarrow`\(x=\dfrac{1}{12}\div3\)
`\Rightarrow`\(x=\dfrac{1}{36}\)
Vậy, `x = 1/36`
`5)`
\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)
`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{52}{21}\)
Vậy, `x = 52/21`
`6)`
\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)
`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(5x=\dfrac{1}{6}\)
`\Rightarrow`\(x=\dfrac{1}{6}\div5\)
`\Rightarrow`\(x=\dfrac{1}{30}\)
Vậy, `x = 1/30.`