\(\left(3x-1\right)^2-1=24\)
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TN
8 tháng 8 2016
A=(x+1)(x+2)(x+3)(x+4)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=(x2+5x+4) ta có:
t(t+2)-24=t2+6t-2t-24
=t(t+6)-4(t+6)
=(t-4)(t+6).Thay vào ta đc:
(x2+5x+4-4)(x2+5x+4+6)=(x2+5x)(x2+5x+10)
=x(x+5)(x2+5x+10)
B=(x2+3x+2)(x2+7x+120-24)
=(x2+3x+2)(x2+7x+96)
=(x2+2x+x+2)(x2+7x+96)
=[x(x+2)+(x+2)](x2+7x+96)
=(x+1)(x+2)(x2+7x+96)
C và D bn cx lm tương tự
BP
3
NH
24 tháng 11 2019
\(\left(2x^2+3x-1\right)^2-5\left(2x^2+3x+3\right)+24=0\)(1)
Đặt \(2x^2+3x+1=a\)
Thay vào (1) ta được \(\left(a-2\right)^2-5\left(a+2\right)+24=0\)
\(\Leftrightarrow a^2-4a+4-5a-10+24=0\)
\(\Leftrightarrow a^2-9a+18=0\)
\(\Leftrightarrow a^2-3a-6a+18=0\)
\(\Leftrightarrow\left(a-3\right)\left(a-6\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3\\a=6\end{matrix}\right.\)
Suy ra \(\left[{}\begin{matrix}2x^2+3x+1=3\\2x^2+3x+1=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+3x-2=0\\2x^2+3x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=0,5\\x=-2\end{matrix}\right.\\\left[{}\begin{matrix}x=1\\x=-2,5\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{0,5;-2,5;1;-2\right\}\)
28 tháng 8 2021
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
TB
1
21 tháng 3 2020
x3 + 3x2 + 3x = 7
<=> x3 + 3x2 + 3x - 7 = 0
<=> (x - 1)(x2 + 4x + 7) = 0
<=> x - 1 = 0 hoặc x2 + 4x + 7 khác 0
<=> x - 1 = 0
<=> x = 1
21 tháng 3 2020
a) ( x2 + 3 x + 2 ) . ( x2 + 3x+ 3 ) - 2 =0
<=>x4 + 3x3 + 3x2 + 3x3 + 9x2 + 9x + 2x2 + 6x + 6 - 2 = 0
<=> x4 + 6x3 + 14x2 + 15x + 4 = 0
<=> x4 + 3x3 + 3x3 + x2 + 9x2 + 4x2 + 3x + 12x + 4 = 0
<=> x2 . ( x2 +3x + 1 ) + 3x . ( x2 +3x + 1 ) + 4. ( x2 + 3x + 1 ) = 0
<=> ( x2 + 3x + 1 ) . ( x2 + 3x + 4 ) = 0
<=> \(\orbr{\begin{cases}x^2+3x+1=0\\x^2+3x+4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)
\(x\notinℝ\)
<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)
Nghiệm cuối cùng là : x1 = \(\frac{-3+\sqrt{5}}{2}\);x2 = \(\frac{-3-\sqrt{5}}{2}\)
b) ( x + 1 ) . ( x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0
<=> ( x2 + 2x + x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0
<=> ( x2 + 3.x + 2 ) . ( x+3) . ( x + 4 ) -24 = 0
<=> ( x3 + 3.x 2 + 3.x2 + 9x + 2x + 6 ) . ( x + 4 ) - 24 = 0
<=> ( x3 + 3x + 2 ) . ( x + 3 ) .( x + 4 ) = 0
<=> ( x3 + 3x2 + 3x2 + 9x + 2x + 6 ) . ( x + 4) - 24 = 0
<=> ( x3 + 6.x2 + 11.x + 6 ) . ( x + 4 ) -24 = 0
<=> x4 + 4.x3 + 6.x3 + 24.x2 + 11.x2 + 44.x + 6.x + 24 - 24 =0
<=> x4 + 10.x3+ 35. x2 + 50.x = 0
<=> x. ( x3 + 10.x2 + 35 .x + 50 ) = 0
<=> x. ( x3 + 5.x2 +5.x2 + 25.x+ 10 + 50 ) = 0
<=> x. [ x2 . ( x+5 ) + 5.x. ( x+5 ) + 10.( x + 5 ) ] = 0
<=> x. ( x + 5 ) . ( x2 + 5.x + 10 ) = 0
=> \(\hept{\begin{cases}x=0\\x+5=0\\x^2+5.x+10=0\end{cases}}\)
=> \(\hept{\begin{cases}x=0\\x=-5\\x\notinℝ\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)
Nghiệm cuối cùng là : x1 = -5 ; x2 = 0
c) x3 + 3.x2 + 3x = 7
<=> x3 + 3.x2 + 3x - 7 = 0
<=> ( x + 1 )3 - 8 = 0
<=> ( x + 1 )3 = 8
<=> ( x + 1 ) 3 = 23
<=> x + 1 = 2
<=> x =1
Vậy x = 1
AT
23 tháng 8 2018
c/ đk: x khác 1; x khác -3
\(\dfrac{3x-1}{x-1}+\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
\(\Rightarrow\left(3x+1\right)\left(x+3\right)+\left(2x+5\right)\left(x-1\right)+4=x^2+2x-3\)
\(\Leftrightarrow3x^2+10x+3+2x^2+3x-5+4=x^2+2x-3\)
\(\Leftrightarrow4x^2+11x+5=0\)
\(\Leftrightarrow\left(4x^2+2\cdot2x\cdot\dfrac{11}{4}+\dfrac{121}{16}\right)-\dfrac{41}{16}=0\)
\(\Leftrightarrow\left(2x+\dfrac{11}{4}\right)^2=\dfrac{41}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{11}{4}=\dfrac{\sqrt{41}}{4}\\2x+\dfrac{11}{4}=-\dfrac{\sqrt{41}}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-11+\sqrt{41}}{8}\\x=\dfrac{-11-\sqrt{41}}{8}\end{matrix}\right.\)
Vậy.........
d/ \(\dfrac{12x+1}{6x-2}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(9x^2-1\right)}\)
đk: \(x\ne\pm\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{12x+1}{2\left(3x-1\right)}-\dfrac{9x-5}{3x+1}=\dfrac{108x-36x^2-9}{4\left(3x-1\right)\left(3x+1\right)}\)
\(\Rightarrow2\left(12x+1\right)\left(3x+1\right)-4\left(9x-5\right)\left(3x-1\right)=108x-36x^2-9\)
\(\Leftrightarrow72x^2+24x+6x+2-108x^2+36x-60x-20-108x+36x^2+9=0\)
\(\Leftrightarrow-102x-9=0\)
\(\Leftrightarrow-102x=9\Leftrightarrow x=-\dfrac{3}{34}\)(TM)
Vậy.........
AT
23 tháng 8 2018
a/ \(\left(x+1\right)^2\left(x+2\right)+\left(x+1\right)^2\left(x-2\right)=-24\)
\(\Leftrightarrow\left(x+1\right)^2\left(x+2+x-2\right)=-24\)
\(\Leftrightarrow2x\left(x^2+2x+1\right)=-24\)
\(\Leftrightarrow2x^3+4x^2+2x+24=0\)
\(\Leftrightarrow2x^3-2x^2+8x+6x^2-6x+24=0\)
\(\Leftrightarrow x\left(2x^2-2x+8\right)+3\left(2x^2-2x+8\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+8\right)\left(x+3\right)=0\)
\(\Leftrightarrow2\left(x^2-x+4\right)\left(x+3\right)=0\)
Ta thấy: \(x^2-x+4=\left(x^2-2x\cdot\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}>0\)
=> x+ 3 = 0 <=> x= -3
Vậy......
b/ \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+x^2+5x+2x^2+x+5=0\)
\(\Leftrightarrow x\left(2x^2+x+5\right)+\left(2x^2+x+5\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
Ta thấy: \(2x^2+x+5=\left(\sqrt{2}x+2\cdot\sqrt{2}x\cdot\dfrac{\sqrt{2}}{4}+\dfrac{1}{8}\right)+\dfrac{39}{8}=\left(\sqrt{2}x+\dfrac{\sqrt{2}}{4}\right)^2+\dfrac{39}{8}>0\)
=> x + 1 = 0 <=> x = -1
Vậy....
\(\left(3x-1\right)^2=25\)
\(\left(3x-1\right)^2=5^2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)
\(\left(3x-1\right)^2=25\)
\(\Leftrightarrow3x-1=5\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=2\)