x67 + 24x= 5761 b) 6x4 + 24 x = 1812
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^3-2x^2+x-2=0\\ \Leftrightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ Vậy:x=2\\ ---\\ 2x\left(3x-5\right)=10-6x\\ \Leftrightarrow6x^2-10x-10+6x=0\\ \Leftrightarrow6x^2-4x-10=0\\ \Leftrightarrow6x^2+6x-10x-10=0\\ \Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(6x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}6x-10=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
\(4-x=2\left(x-4\right)^2\\ \Leftrightarrow4-x=2\left(x^2-8x+16\right)\\ \Leftrightarrow2x^2-16x+32+x-4=0\\ \Leftrightarrow2x^2-15x+28=0\\ \Leftrightarrow2x^2-8x-7x+28=0\\ \Leftrightarrow2x\left(x-4\right)-7\left(x-4\right)=0\\ \Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\\ ---\\ 4-6x+x\left(3x-2\right)=0\\ \Leftrightarrow4-6x+3x^2-2x=0\\ \Leftrightarrow3x^2-8x+4=0\\ \Leftrightarrow3x^2-6x-2x+4=0\\ \Leftrightarrow3x\left(x-2\right)-2\left(x-2\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
a) =x3-2x2+6x2-12x -12x +24
= x2(x-2)+6x(x-2)-12(x-2)
= (x-2)(x2+6x-12)
mk giải đc câu a thôi, bn zô jup mk lại vs
\(a,x^3+4x^2-24x+24\)
\(=x^3+6x^2-12x-2x^2-12x+24\)
\(=\left(x^3-2x^2\right)+\left(6x^2-12x\right)-\left(12x-24\right)\)
\(=x^2\left(x-2\right)+6x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+6x-12\right)\)
(x2 + 5x + 6)(x2 + 9x + 20) = 24
<=> (x + 2)(x + 3)(x + 4)(x + 5) - 24 = 0
<=> (x2 + 7x + 10)(x2 + 7x + 12) - 24 = 0 (1)
Đặt x2 + 7x + 11 = t, ta có:
(1) <=> (t - 1)(t + 1) - 24 = 0
<=> t2 - 1 - 24 = 0
<=> (t - 5)(t + 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}t-5=0\\t+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+7x+11-5=0\\x^2+7x+11+5=0\end{matrix}\right.\)
<=> (x + 1)(x + 6) = 0 (vì \(x^2+7x+16\ge\dfrac{15}{4}>0\))
<=> x = - 1 hoặc x = - 6
~ ~ ~ ~ ~
x4 - 24x = 32
<=> x4 - 24x - 32 = 0
<=> (x2 - 2x - 4)(x2 + 2x + 8) = 0
<=> \(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)=0\) (vì \(x^2+2x+8\ge7>0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{matrix}\right.\)
\(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(=24^3-3\cdot24\cdot18\)
\(=13824-1296\)
=12528
24.(16-5)-16.(24-5)=24.16-24.5-16.24+16.5
=(24.16-24.16)-5.(24-16)
=-5.8=-40
Giải luôn nhé!
= 24 x 16 - 5 - 16 x 24 - 5
= 24 x 11 - 24 x 16 - 5
= 24 x 11 - 24 x 11
= 0