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TN
1
TN
11 tháng 3 2016
<=>(4x-3)3+(5-7x)3+(3x-8)3=-3(3x-8)(4x+3)(7x-5)
=>-3(3x-8)(4x+3)(7x-5)=0
Th1:-3(3x-8)=0
=>3x-8=0
=>3x=8
=>x=\(\frac{8}{3}\)
Th2:4x+3=0
=>4x=-3
=>x=\(-\frac{3}{4}\)
Th3:7x-5=0
=>7x=5
=x=\(\frac{5}{7}\)
NT
2
29 tháng 10 2017
Đặt \(\left(4x+3\right)^3=a^3\\ \left(5-7x\right)^3=b^3\\ \left(3x-8\right)^3=c^3\)
Từ giả thuyết ta có:
\(a^3-b^3+c^3=0\\ \Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b\right)=0\\ \Leftrightarrow\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b\right)=0\\ \Leftrightarrow\left(4x+3+5-7x+3x-8\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b\right)=0\\ \Leftrightarrow0\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b\right)=0\\ \Leftrightarrow-3ab\left(a+b\right)=0.\)
LM
0
TQ
4 tháng 3 2018
\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow\left(2x^3+7x^2+7x\right)+2=0\)
\(\Leftrightarrow x\left(2x^2+7x+7+2\right)=0\)
\(\Leftrightarrow x\left(2x^2+7x+9\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x+3x+9\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2+6x\right)+\left(3x+9\right)\right]=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=-\dfrac{3}{2}\end{matrix}\right.\)
chúc bạn học tốt!
28 tháng 6 2021
a) (7x - 8)(7x + 8) - 10(2x + 3)2 + 5x(3x - 2)2 - 4x(x - 5)2
= 49x2 - 64 - 10(4x2 + 12x + 9) + 5x(9x2 - 12x + 4) - 4x(x2 - 10x + 25)
= 49x2 - 64 - 40x2 - 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x - 100x
= 41x3 - 51x2 - 160x - 154
b) (x2 - 3)(x2 + 3) - 5x2(x + 1)2 - (x2 - 3x)(x2 - 2x) + 4x(x + 2)2
= x4 - 9 - 5x2(x2 + 2x + 1) - x4 + 5x3 - 6x2 + 4x(x2 + 4x + 4)
= 5x3 - 6x2 - 5x4 - 10x3 - 5x2 + 4x3 + 16x2 + 16x - 9
= -5x4 - x3 + 5x2 + 16x - 9
QA
28 tháng 6 2021
Trả lời:
a , ( 7x - 8 ) ( 7x + 8 ) - 10 ( 2x + 3 )2 + 5x ( 3x - 2 )2 - 4x ( x - 5 )2
= 49x2 - 64 - 10 ( 4x2 + 12x + 9 ) + 5x ( 9x2 - 12x + 4 ) - 4x ( x2 - 10x + 25 )
= 49x2 - 64 - 40x2 + 120x - 90 + 45x3 - 60x2 + 20x - 4x3 + 40x2 - 100x
= 41x3 - 11x2 + 40x - 154
b , ( x2 - 3 ) ( x2 + 3 ) - 5x2 ( x + 1 )2 - ( x2 - 3x ) ( x2 - 2x ) + 4x ( x + 2 )2
= x4 - 9 - 5x2 ( x2 + 2x + 1 ) - ( x4 - 2x3 - 3x3 + 6x2 ) + 4x ( x2 + 4x + 4 )
= x4 - 9 - 5x4 - 10x3 - 5x2 - x4 + 2x3 + 3x3 - 6x2 + 4x3 + 16x2 + 16x
= - 5x4 - x3 + 5x2 + 16x - 9
6 tháng 7 2018
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
3 tháng 3 2019
a/ Đặt : \(\left\{{}\begin{matrix}a=4x-3\\b=3x-2\end{matrix}\right.\) \(\Leftrightarrow a+b=7x-5\)
Thay vào pt ta dc :
\(a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow a^3+b^3=a^3+3a^2b+3ab^2+b^3\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)
+) \(a=0\Leftrightarrow4x-3=0\Leftrightarrow x=\dfrac{3}{4}\)
+) \(b=0\Leftrightarrow3x-2=0\Leftrightarrow x=\dfrac{2}{3}\)
+) \(c=0\Leftrightarrow7x-3=0\Leftrightarrow x=\dfrac{3}{7}\)
Vậy...
b/ \(x^3-2x^2-x-6=0\)
\(\Leftrightarrow x^3-3x^2+x^2-3x+2x-6=0\)
\(\Leftrightarrow x^2\left(x-3\right)+x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right]=0\)
Mà \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy..
3 tháng 3 2019
a) (4x - 3)3 + (3x - 2)3 = (7x - 5)3
\(\Leftrightarrow\) (4x - 3)3 + (3x - 2)3 = (4x - 3)3 + (3x - 2)3 + 3(4x - 3)(3x - 2)(4x - 3 + 3x - 2)
\(\Leftrightarrow\) 3(4x - 3)(3x - 2)(7x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{3}\\x=\dfrac{5}{7}\end{matrix}\right.\)
DN
5 tháng 3 2019
\(j,3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy...............................
DN
5 tháng 3 2019
\(m,3x^2+4x-4=0\)
\(\Leftrightarrow3x^2+6x-2x-4=0\)
\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)
Áp dụng hằng đẳng thức :
\(a^3+b^3+c^3=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)
Ta viết lại phương trình thành :
\(\left(4x+3+5-7x+3x-8\right)\left[\left(4x+3\right)^2+\left(5-7x\right)^2+\left(3x-8\right)^2\right]+3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)
\(\Leftrightarrow3\left(4x+3\right)\left(5-7x\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\5-7x=0\\3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{7}\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy : \(S=\left\{-\dfrac{3}{4};\dfrac{5}{7};\dfrac{8}{3}\right\}\).