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\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

\(\Leftrightarrow x^2\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

2 tháng 5 2023

Vì chia hết cho 5 

\(\Rightarrow y=0\) hoặc \(y=5\)

\(Th1:y=0\\ \Rightarrow x-0=6\\\Rightarrow x=6\)           \(\Rightarrow x=6;y=0\)

\(Th2:y=5\\ \Rightarrow x-5=6\\ \Rightarrow x=11\)           \(\Rightarrow x=11;y=5\)

\(\Rightarrow A\)

12 tháng 12 2021

Answer:

\(3x^2-4x=0\)

\(\Rightarrow x\left(3x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)

\(\left(x^2-5x\right)+x-5=0\)

\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

\(x^2-5x+6=0\)

\(\Rightarrow x^2-2x-3x+6=0\)

\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)

\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(5x\left(x-3\right)-x+3=0\)

\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)

\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)

\(x^2-2x+5=0\)

\(\Rightarrow\left(x^2-2x+1\right)+4=0\)

\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)

Vậy không có giá trị \(x\) thoả mãn

\(x^2+x-6=0\)

\(\Rightarrow x^2+3x-2x-6=0\)

\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)

AH
Akai Haruma
Giáo viên
11 tháng 12 2023

Lời giải:

a. $x(\frac{4}{5}x-1)(0,1x-10)=0$

$\Rightarrow x=0$ hoặc $\frac{4}{5}x-1=0$ hoặc $0,1x-10=0$

Nếu $\frac{4}{5}x-1=0$

$\Rightarrow x=1: \frac{4}{5}=\frac{5}{4}$

Nếu $0,1x-10=0$

$\Rightarrow x=10:0,1=100$

Vậy $x=0; \frac{5}{4}; 100$

b.

$(\frac{1}{4}x-1)-(\frac{5}{6}x+2)-(1-\frac{5}{8}x)=0$

$(\frac{1}{4}x-\frac{5}{6}x+\frac{5}{8}x)-(1+2+1)=0$

$\frac{1}{24}x-4=0$

$x=4: \frac{1}{24}=96$

27 tháng 12 2016

5+5+8+0+6+4+5+2+4+1+1+2+3+4+5+6+6+7+8+9+100000000+45638+78536 x 12345 x 34 x 0 +100=100045829

CHÚC BẠN HỌC GIỎI

TK MÌNH NHÉ

27 tháng 12 2016

Dạ, cảm ơn anh, nhưng đây kg phải là đáp án của em. TRẻ con còn biết, mà người lại kg .Anh đừng tự tin, nhanh rồi cũng có lúc sai!!!

3 tháng 11 2018

\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)

\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)

\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)

\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)

\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)

1 tháng 3 2020

1. \(\Leftrightarrow\left(x-6\right)\left(x+7\right)+5\left(x-6\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left[\left(x+7\right)+5\left(3x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-6\right)\left(16x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\16x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-\frac{1}{8}\end{matrix}\right.\)

1 tháng 3 2020

4. \(\Leftrightarrow\left(x+5\right)^2\left(3x+2\right)^2-x^2\left(x+5\right)^2=0\)

\(\Leftrightarrow\left(x+5\right)^2\left[\left(3x+2\right)^2-x^2\right]=0\)

\(\Leftrightarrow\left(x+5\right)^2\left(2x+2\right)\left(4x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=0\\2x+2=0\\4x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x=-2\\4x=-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-1\\x=-\frac{1}{2}\end{matrix}\right.\)

2 tháng 1 2019

6 x 1 = 6              6 x 9 = 54

6 x 2 = 12              6 x 8 = 48

6 x 3 = 18              6 x 7 = 42

6 x 4 = 24              0 x 6 = 0

6 x 5 = 30              6 x 0 = 0

6 x 6 = 36              6 x 10 = 60

16 tháng 9 2018

6 x 4 =24

6 x 1 =6

6 x 9 =54

6 x 10 =60

6 x 6 =36

6 x 3 =18

6 x 2 =12

0 x 6 =0

6 x 8 =48

6 x 5 =30

6 x 7 =42

6 x 0 =0

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)