1/36-(2-3x)^2=0
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Y
7 tháng 7 2023
\(5,4x^2-36=0\\ \Leftrightarrow\left(2x\right)^2-6^2=0\\ \Leftrightarrow\left(2x-6\right)\left(2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy \(S=\left\{3;-3\right\}\)
\(7,\left(3x+1\right)^2-16=0\\ \Leftrightarrow\left(3x+1\right)^2-4^2=0\\ \Leftrightarrow\left(3x+1-4\right)\left(3x+1+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-3=0\\3x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{5}{3}\right\}\)
\(8,\left(2x-3\right)^2-49=0\\ \Leftrightarrow\left(2x-3\right)^2-7^2=0\\ \Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-10=0\\2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{-2;5\right\}\)
2 tháng 3 2020
a, Ta có : \(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\)
=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
=> \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
=> \(\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)
=> \(\left(x-1\right)\left(4x-3\right)=0\)
=> \(\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{1,\frac{3}{4}\right\}\)
b, Ta có : \(5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\)
=> \(5x^2+15x-27x-27=x^2-36\)
=> \(5x^2+15x-27x-27-x^2+36=0\)
=> \(4x^2-12x+9=0\)
=> \(\left(2x-3\right)^2=0\)
=> \(x=\frac{3}{2}\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{3}{2}\right\}\)
TN
2 tháng 3 2020
\(a.\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{1;\frac{3}{4}\right\}\)
\(b.5\left(x^2+3x\right)-9\left(3x+3\right)=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27=x^2-36\\ \Leftrightarrow5x^2+15x-27x-27-x^2+36=0\\ \Leftrightarrow4x^2-12x+9=0\\ \Leftrightarrow\left(2x-3\right)^2=0\\ \Leftrightarrow x=\frac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\frac{3}{2}\right\}\)
Chúc bạn học tốt!!!!!!!!!!!
10 tháng 12 2021
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
Bài 1.
1 tháng 2 2017
\(\left(x+6\right)\left(3x+1\right)+x^2-36=0\)
\(\Leftrightarrow\left(x+6\right)\left(3x+1\right)+\left(x-6\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(3x+1+x-6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+6=0\\4x-5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-6\\x=\frac{5}{4}\end{cases}}\)
Vậy \(x=-6;\frac{5}{4}\)
11 tháng 3 2020
a) (x + 6)(3x + 1) + x2 - 36 = 0
<=> 3x2 + x + 18x + 6 + x2 - 36 = 0
<=> 4x2 + 19x - 30 = 0
<=> 4x2 + 24x - 5x - 30 = 0
<=> 4x(x + 6) - 5(x + 6) = 0
<=> (x + 6)(4x - 5) = 0
<=> x + 6 = 0 hoặc 4x - 5 = 0
<=> x = -6 hoặc x = 5/4
11 tháng 3 2020
Bài 1 mình đã làm xong rồi, anh em nào giúp mình bài 2 với!
6 tháng 2 2020
Bài 1 :
a, \(\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
=> \(\left(x-3\right)\left(4x-1-5x-2\right)=0\)
=> \(\left(x-3\right)\left(-x-3\right)=0\)
=> \(\left[{}\begin{matrix}x-3=0\\-x-3=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=\pm3\) .
b, \(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)
=> \(\left(x+3\right)\left(x-5+3x-4\right)=0\)
=> \(\left(x+3\right)\left(4x-9\right)=0\)
=> \(\left[{}\begin{matrix}x+3=0\\4x-9=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-3\\x=\frac{9}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-3,x=\frac{9}{4}\) .
c, \(\left(x+6\right)\left(3x-1\right)+x^2-36=0\)
=> \(\left(x+6\right)\left(3x-1\right)+\left(x-6\right)\left(x+6\right)=0\)
=> \(\left(x+6\right)\left(3x-1+x-6\right)=0\)
=> \(\left(x+6\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=-6,x=\frac{7}{4}\) .
6 tháng 2 2020
a) ( 4x - 1 ) ( x - 3 ) - ( x - 3 ) ( 5x + 2 ) = 0
⇔ ( x - 3 ) ( 4x - 1 - 5x - 2 ) = 0
⇔ ( x - 3 ) ( -x - 3 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Ý b) tương tự ý a) thôi.
c) ( x + 6 ) ( 3x - 1 ) + x2 - 36 = 0
⇔ ( x + 6 ) ( 3x - 1 ) + ( x + 6 ) ( x - 6 ) = 0
⇔ (x+6)(3x-1+x-6)=0
⇔ (x+6)(4x-7)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\4x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\frac{7}{4}\end{matrix}\right.\)
HT
2
14 tháng 2 2015
a/ (4x-1)(x-3)-(x-3)(5x+2)=0
<=> (x-3)(4x-1-5x-2)=0
<=> (x-3)(-x-3)=0
<=> x-3=0 hoặc -x-3=0
<=> x=3 hoặc x= -3
b/ (x+6)(3x-1)+ x^2 -36 =0
<=> (x+6)(3x-1) + (x-6)(x+6)=0
<=> (x+6)(3x-1+x-6)=0
<=> (x+6)(4x-7)=0
<=> x+6=o hoặc 4x-7=0
<=> x= -6 hoặc x= 7/4
c/ (x+3)(x+5)+(x+3)(3x-4)=0
<=> (x+3)(x+5+3x-4)=0
<=> (x+3)(4x+1)=0
<=> x+3=0 hoặc 4x+1=0
<=> x= -3 hoặc x=-1/4
\(\left(2-3x\right)^2=\dfrac{1}{36}=\left(\dfrac{1}{6}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2-3x=\dfrac{1}{6}\\2-3x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{18}\\x=\dfrac{13}{18}\end{matrix}\right.\)
\(\left[{}\begin{matrix}2-3x=\dfrac{1}{6}\\2-3x=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{11}{6}\\3x=\dfrac{13}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{6}:3=\dfrac{11}{18}\\x=\dfrac{13}{6}:3=\dfrac{13}{18}\end{matrix}\right.\)