Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài này đề bài phải là khai triển biểu thức, chứ không phải là tính em nhé.
Lời giải:
Ta áp dụng hằng đẳng thức đáng nhớ thôi.
a. $(3+2x)^3=3^3+3.3^2.2x+3.3.(2x)^2+(2x)^3$
$=8x^3+36x^2+54x+27$
b.
$(\frac{1}{2}-y)^3=(\frac{1}{2})^3-3.(\frac{1}{2})^2.y+3.\frac{1}{2}y^2-y^3$
$=-y^3+\frac{3}{2}y^2-\frac{3}{4}y+\frac{1}{8}$
c.
$(x-5)(x^2+5x+25)=(x-5)^2(x^2+5x+5^2)$
$=x^3-5^3=x^3-125$
d.
$(3x+\frac{1}{2})(9x^2-\frac{3}{2}x+\frac{1}{4})$
$=(3x+\frac{1}{2})[(3x)^2-3x.\frac{1}{2}+(\frac{1}{2})^2]$
$=(3x)^3+(\frac{1}{2})^3=27x^3+\frac{1}{8}$
a) \(6x^2-15x\)
b) \(x^2+5x+4\)
c) \(49-x^2\)
d) \(x^2+4x+4\)
e) \(9-12x+4x^2\)
f) \(x^3-8\)
\(a,=6x^2-15x\\ b,=x^2+5x+4\\ c,=49-x^2\\ d,=x^2+4x+4\\ e,=9-12x+4x^2\\ f,=x^3-8\)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
a. 6x2 - (2x + 5)(3x - 2) = 7
<=> 6x2 - 6x2 + 4x - 15x + 10 = 7
<=> -11x = -3
<=> \(x=\dfrac{3}{11}\)
b. (5 - x)(25 + 5x + x2) + x(x2 - 7) = 25
<=> 125 - x3 + x3 - 7x = 25
<=> -7x = 25 - 125
<=> -7x = -100
<=> \(x=\dfrac{100}{7}\)
c. (7 - 2x)2 + (3 + 2x)(3 - 2x) = 30
<=> 49 - 28x + 4x2 + 9 - 4x2 = 30
<=> 4x2 - 4x2 - 28x = 30 - 49 - 9
<=> -28x = -28
<=> x = 1
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
\(\left\{{}\begin{matrix}\left\{{}\begin{matrix}3x-15=45\Leftrightarrow3x=60\Leftrightarrow x=20\\35-5x=50\Leftrightarrow5x=-15\Leftrightarrow x=-3\end{matrix}\right.\\\left\{{}\begin{matrix}\left(2x-5\right)+17=6\Leftrightarrow2x+5=-11\Leftrightarrow2x=-16\Leftrightarrow x=-8\\10-2\left(4-3x\right)=-4\Leftrightarrow8-6x=14\Leftrightarrow6x=-6\Leftrightarrow x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}-12+3\left(-x+7\right)=-18\Leftrightarrow-3x+21=-6\Leftrightarrow-3x=-27\Leftrightarrow x=9\\24:\left(3x-2\right)=-3\Leftrightarrow3x-2=-8\Leftrightarrow3x=-6\Leftrightarrow x=-2\end{matrix}\right.\\-45:5\left(-3-2x\right)=3\Leftrightarrow-15-10x=-15\Leftrightarrow10x=0\Leftrightarrow x=0\end{matrix}\right.\)
Bài 1.
a)\(\left(-0,125\right)^5\cdot\left(-8\right)^5=\left(\left(-0,125\right)\cdot\left(-8\right)\right)^5=1^5=1\)