ko phải "-" đâu mà là nhân đấy

giữa 2 ngoặc đó

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a, \(6x\left(x-5\right)-\left(2x+7\right)\left(3x-2\right)=0\)

\(\Rightarrow6x^2-30x-\left(6x^2-4x+21x-14\right)=0\)

\(\Rightarrow6x^2-30x-6x^2+4x-21x+14=0\)

\(\Rightarrow-47x=-14\Rightarrow x=\dfrac{14}{47}\)

b, \(\left(8x+3\right)x-\left(4x+1\right)\left(2x-7\right)=5\)

\(\Rightarrow8x^2+3x-\left(8x^2-28x+2x-7\right)=5\)

\(\Rightarrow8x^2+3x-8x^2+28x-2x+7=5\)

\(\Rightarrow29x=5-7\Rightarrow29x=-2\)

\(\Rightarrow x=\dfrac{-2}{29}\)

c, \(\left(2x-3\right)\left(2x+3\right)-x.\left(4x+1\right)=1\)

\(\Rightarrow\left(2x\right)^2-9-4x^2-x=1\)

\(\Rightarrow4x^2-4x^2-x=1+9\)

\(\Rightarrow-x=10\Rightarrow x=-10\)

Chúc bạn học tốt!!!

a ) \(6x\left(x-5\right)-\left(2x+7\right)\left(3x-2\right)=0\)

\(\Leftrightarrow6x^2-30x-\left(6x^2-4x+21x-14\right)=0\)

\(\Leftrightarrow6x^2-30x-6x^2+4x-21x+14=0\)

\(\Leftrightarrow-47x+14=0\)

\(\Leftrightarrow x=\dfrac{14}{47}.\)

Vậy ..........................................

b ) \(\left(8x+3\right).x-\left(4x+1\right)\left(2x-7\right)=5\)

\(\Leftrightarrow8x^2+3x-\left(8x^2-28x+2x-7\right)=5\)

\(\Leftrightarrow8x^2+3x-8x^2+28x-2x+7=5\)

\(\Leftrightarrow29x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{29}.\)

Vậy.............

c ) \(\left(2x-3\right)\left(2x+3\right)-x\left(4x+1\right)=1\)

\(\Leftrightarrow4x^2-9-4x^2-x=1\)

\(\Leftrightarrow-x=10\)

\(\Leftrightarrow x=10\)

Vậy..........

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

a, \(=2x^2-2x\)

b, \(=-4x^2-4x+3\)

a) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt a = \(8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

mà \(8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\)

\(\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-0,25\end{cases}}\)

cảm ơn bạn còn mấy phần còn lại ạ

\(a.x=-0,6\)

\(c.x=-11,6\)

Pt nhju ak!!!

\(\dfrac{x-1}{2x^2-4x}-\dfrac{7}{8x}=\dfrac{5-x}{4x^2-8x}-\dfrac{1}{8x-16}\) ( ĐKXĐ: \(x\ne0;x\ne2\) )

\(\Leftrightarrow\dfrac{x-1}{2x\left(x-2\right)}-\dfrac{7}{8x}=\dfrac{5-x}{4x\left(x-2\right)}-\dfrac{1}{8\left(x-2\right)}\)

\(\Leftrightarrow\dfrac{\left(x-1\right)4}{8x\left(x-2\right)}-\dfrac{7\left(x-2\right)}{8x\left(x-2\right)}=\dfrac{2\left(5-x\right)}{8x\left(x-2\right)}-\dfrac{1x}{8x\left(x-2\right)}\)

\(\Rightarrow4x-4-7x+14=10-2x-x\)

\(\Leftrightarrow-3x+2x+x=10+4-14\)

\(\Leftrightarrow0=0\)

          Vậy pt đã cho có nghiệm đúng với mọi x

\(\left|2+3x\right|=\left|4x-3\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{7};5\right\}\)

\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)

\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

Giải tiếp tương tự

Sau đó giải tiếp câu còn lại

a, \(\left|2+3x\right|=\left|4x-3\right|\)

\(\Rightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x+3\end{cases}\Rightarrow\orbr{\begin{cases}3x-4x=-3-2\\3x+4x=3-2\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-5\\7x=1\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}}}\)

Câu b tương tự

c, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

\(\Rightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}\frac{5}{4}x-\frac{5}{8}x=\frac{3}{5}+\frac{7}{2}\\\frac{5}{4}x+\frac{5}{8}x=-\frac{3}{5}+\frac{7}{2}\end{cases}\Rightarrow}\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}}\)

d, \(\left|\frac{7}{5}x+\frac{3}{2}\right|-\left|\frac{4}{3}-\frac{1}{4}\right|=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|-\frac{13}{12}=0\)

\(\Rightarrow\left|\frac{7}{5}x+\frac{3}{2}\right|=\frac{13}{12}\)

Đến đây dễ rồi, tự làm tiếp :)

P/s: Ko chắc, sai ib với t :v