Giải phương trình: \((x+2)(x+3)(x+8)(x+12)\)
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LD
1
D
7 tháng 9 2019
⇔ [( x + 2 )( x+12 )][( x + 3 )(x + 8)] = 4x²
⇔ ( x\(^2\) + 2x + 12x + 24 ) ( x\(^2\) + 3x + 8x + 24 ) = 4x²
Đặt x\(^2\) + 24 là a tacó :
pt⇔( a + 14x )( a + 11x ) = 4x\(^2\)
⇔ a\(^2\) + 11ax + 14ax + 154x\(^2\) - 4x\(^2\) = 0
⇔ a\(^2\) + 25ax + 150x\(^2\) = 0
⇔ a\(^2\) + 15ax + 10ax + 150x\(^2\) = 0
⇔ a( a + 15x ) + 10x ( a + 15x ) = 0
⇔ ( a + 10x ) ( a + 15x ) = 0
Thay a bằng x\(^2\) + 24
pt⇔ ( x\(^2\) + 24 + 10x ) ( x\(^2\) + 24 + 15x ) = 0
⇔ ( x\(^2\) + 4x + 6x + 24 ) ( x\(^2\) + 15x + 24 ) = 0
⇔ [ x( x + 4 ) + 6 (x + 4 )] ( np in dam) = 0
⇔ [ ( x + 6 ) ( x + 4 ) ] ( cnt ) = 0
⇔ \(\left[{}\begin{matrix}x+6=0\\x+4=0\\x^2+15x+24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-4\\x\approx-1,82\\x\approx-13,18\end{matrix}\right.\)
DT
2
5 tháng 4 2021
|x-9|=2x+5
Xét 3 TH
TH1: x>9 => x-9=2x+5 =>-9-5=x =>x=-14 (L)
TH2: x<9 => 9-x=2x+5 => 9-5=3x =>x=4/3(t/m)
TH3: x=9 =>0=23(L)
Vậy x= 4/3
5 tháng 4 2021
Ta có:\(\dfrac{1-2x}{4}-2\le\dfrac{1-5x}{8}+x\\ \)
\(\dfrac{2-4x-16}{8}\le\dfrac{1-5x+8x}{8}\)
\(-4x-14\le1+3x\\ \Leftrightarrow7x+15\ge0\\ \Leftrightarrow x\ge-\dfrac{15}{7}\)
NM
0
6 tháng 2 2017
|x-8|+|x+8|= x^2 - 16
= ( x-8)+ ( x+ 8) = x^2 - 16
=> 2x = x^2 - 16
=> đến đây bn tự giải nhé
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 1 2021
\(3x^3+6x^2-12x+8=0\)
\(\Leftrightarrow4x^3=x^3-6x^2+12x-8\)
\(\Leftrightarrow4x^3=\left(x-2\right)^3\)
\(\Rightarrow\sqrt[3]{4}.x=x-2\)
\(\Rightarrow x=\dfrac{2}{1-\sqrt[3]{4}}\)
NM
0
E
1
ND
2 tháng 7 2018
\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{8}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{y}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+3\left(\dfrac{1}{8}-\dfrac{1}{y}\right)=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{4}{x}+\dfrac{3}{8}-\dfrac{3}{x}=\dfrac{5}{12}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\\dfrac{1}{x}=\dfrac{1}{24}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{8}-\dfrac{1}{x}\\x=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{y}=\dfrac{1}{12}\\x=24\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=12\\x=24\end{matrix}\right.\)
1 tháng 1 2018
\(\text{a) }\left(x^2+x\right)^2+4\left(x^2+x\right)=12\\ \Leftrightarrow\text{Đặt }x^2+x=y\\ \Leftrightarrow y^2+4y=12\\ \Leftrightarrow y^2+6y-2y-12=0\\ \Leftrightarrow\left(y^2+6y\right)-\left(2y+12\right)=0\\ \Leftrightarrow y\left(y+6\right)-2\left(y+6\right)=0\\ \Leftrightarrow\left(y+6\right)\left(y-2\right)=0\\ \Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\\ \Leftrightarrow\left(x^2+x+\dfrac{1}{4}+\dfrac{23}{4}\right)\left(x^2+2x-x-2\right)=0\\ \Leftrightarrow\left[\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{23}{4}\right]\left[\left(x^2+2x\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]\left[x\left(x+2\right)-\left(x+2\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\left(Vì\text{ }\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\\ \text{Vậy }S=\left\{1;-2\right\}\\ \)
\(\text{b) }6x^4-5x^3-38x^2-5x+6=0\\ \Leftrightarrow x^2\left(6x^2-5x-38-\dfrac{5}{x}+\dfrac{6}{x^2}\right)=0\\ \Leftrightarrow x^2\left[\left(6x^2+12+\dfrac{6}{x^2}\right)-\left(5x+\dfrac{5}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x^2+2+\dfrac{1}{x^2}\right)-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \Leftrightarrow x^2\left[6\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-50\right]=0\\ \text{Đặt }x+\dfrac{1}{x}=y\\ \Leftrightarrow x^2\left(6y^2-5y-50\right)=0\\ \Leftrightarrow x^2\left(6y^2-20y+15y-50\right)=0\\ \Leftrightarrow x^2\left[\left(6y^2-20y\right)+\left(15y-50\right)\right]=0\\ \Leftrightarrow x^2\left[2y\left(3y-10\right)+5\left(3y-10\right)\right]=0\\ \Leftrightarrow x^2\left(2y+5\right)\left(3y-10\right)=0\\ \Leftrightarrow x^2\left(2x+\dfrac{2}{x}+5\right)\left(3x+\dfrac{3}{x}-10\right)=0\\ \Leftrightarrow\left(2x^2+2+5x\right)\left(3x^2+3-10x\right)=0\\ \Leftrightarrow\left(2x^2+4x+x+2\right)\left(3x^2-9x-x+3\right)=0\\ \Leftrightarrow\left[\left(2x^2+4x\right)+\left(x+2\right)\right]\left[\left(3x^2-9x\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left[2x\left(x+2\right)+\left(x+2\right)\right]\left[3x\left(x-3\right)-\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x+1\right)\left(x+2\right)\left(3x-1\right)\left(x-3\right)=0\\ \)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=-2\\3x=1\\x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\\x=\dfrac{1}{3}\\x=3\end{matrix}\right.\\ \text{Vậy }S=\left\{-\dfrac{1}{2};-2;\dfrac{1}{3};3\right\}\)
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= 4x2 nữa nhé. Mình quên , hihi