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9 tháng 8 2016
Ta có: 3(x^2 -16/3x) +12 = 3((x -16/6)^2-64/9)+12=3(x-16/6)^2 -64/3+12=3(x-16/6)^2 -28/3=0 =>3(x-16/6)^2 =28/3 => (x-16/6)^2 =28/9
=> x-16/6 = 2 căn 7 phần 3 => x= ..
22 tháng 1 2019
a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)
b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x\left(x-4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;4\right\}\)
c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)
\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)
d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)
e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)
g. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)
3 tháng 3 2020
a) (2x + 5)(x - 3) = (x - 4)(3 - x)
<=> (2x + 5)(x - 3) + (x - 3)(x - 4) = 0
<=> (2x + 5 + x - 4)(x - 3) = 0
<=> (3x + 1)(x - 3) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\frac{1}{3}\\x=3\end{matrix}\right.\)
Vậy S = {-1/3; 3}
b) 18x2(x + 4) - 12(x2 + 4x) = 0
<=> 18x2(x + 4) - 12x(x + 4) = 0
<=> 6x(x + 4)(3x - 2) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+4=0\\3x-2=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy S = {0; -2; 2/3}
24 tháng 2 2020
\(4+2x\left(2x+4\right)=-x\)
\(4+2x.2x+8x=-x\)
\(4x+8x+x=-4\)
\(13x=-4\)
\(x=-\frac{4}{13}\)
Vậy pt có nghiệm là { -4/13 }
8 tháng 1 2020
a) 9x2 - 1 = (3x + 1)(2x - 3)
=> 9x2 - 1 = 6x2 - 9x + 2x - 3
=> 9x2 - 6x2 + 7x - 1 + 3 = 0
=> 3x2 + 7x + 2 = 0
=> 3x2 + 6x + x + 2 = 0
=> 3x(x + 2) + (x + 2) = 0
=> (3x + 1)(x + 2) = 0
=>\(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)
b) 2(9x2 + 6x + 1) = (3x + 1)(x - 2)
=> 2(3x + 1)2 - (3x + 1)(x - 2) = 0
=> (3x + 1)(6x + 2 - x + 2) = 0
=> (3x + 1)(5x +4 ) = 0
=> \(\orbr{\begin{cases}3x+1=0\\5x+4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-\frac{4}{5}\end{cases}}\)
c) 27x2(x + 3) - 12(x2 + 3x) = 0
=> 27x2(x + 3) - 12x(x + 3) = 0
=> 3x(9x - 4)(x + 3) = 0
=> 3x = 0
9x - 4 = 0
x + 3 = 0
=> x = 0
x = 4/9
x = -3
d) 16x2 - 8x + 1 = 4(x + 3)(4x - 1)
=> (4x - 1)2 - 4(x + 3)(4x - 1) = 0
=> (4x - 1)(4x - 1 - 4x - 12) = 0
=> 4x - 1 = 0
=> x = 1/4
TU
0
MA
1
AH
Akai Haruma
Giáo viên
16 tháng 12 2023
Lời giải:
$(4x-2)^2-16x+16x=(4x-2)^2$
----------------
$(3x-4)(3x+4)-(20x^2y-15xy^2):(5xy)$
$=(3x-4)(3x+4)-(4x-3y)$ không phân tích được thành nhân tử.
-----------------------------------
$(x-2)(3x^2+6x+12)-(120^2x^2y^2):(60xy^2)$
$=3(x-2)(x^2+2x+4)-240x$
$=3(x^3-2^3)-240x=3x^3-240x-24$
$=3(x^3-80x-8)$
3x^ 2 - 16x- 12 = 0
<=>3x2+2x-18x-12=0
<=>x(3x+2)-6.(3x+2)=0
<=>(3x+2)(x-6)=0
<=>3x+2=0 hoặc x-6=0
<=>3x=-2 hoặc x=6
<=>x=-2/3 hoặc x=6