b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

\(a,\left(x-8\right)\left(x^3+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

\(b,\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\\ \Leftrightarrow4x-3-x-5=30-3x\\ \Leftrightarrow3x-8-30+3x=0\\ \Leftrightarrow6x-38=0\\ \Leftrightarrow x=\dfrac{19}{3}\)

TK

`a.(x-8)(x+8)=0`

`⇔³{x−8=0x³+8=2 `

`⇔³³{x=8x³=−2³ `

`⇔{x=8x=−2`

Vậy ` x = 8;-2`

`b. ( 4 x − 3 ) − ( x + 5 ) = 3 . ( 10 − x )`

`⇔ 4 x − 3 − x − 5 = 30 − 3 x`

`⇔ 3 x − 8 = 30 − 3 x`

`⇔ 3 x + 3 x = 30 + 8`

`⇔ 6 x = 38`

`⇔ x = 19/ 3`

Vậy ` x = 19/ 3`

đề thiếu thì ai mà làm hộ được chứ 

không giải được, bài thiếu đề

d: \(\frac{3}{x}+\frac{y}{3}=\frac56\)

=>\(\frac{3}{x}=\frac56-\frac{y}{3}=\frac{5-2y}{6}\)

=>x(5-2y)=18

=>x(2y-5)=-18

mà 2y-5 lẻ và x>0

nên (x;2y-5)∈{(18;1);(2;9);(6;3)}

=>(x;2y)∈{(18;6);(2;14);(6;8)}

=>(x;y)∈{(18;3);(2;7);(6;4)}

c: \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)

=>\(\frac{x}{9}-\frac{1}{18}=\frac{3}{y}\)

=>\(\frac{2x-1}{18}=\frac{3}{y}\)

=>y(2x-1)=54

mà 2x-1 lẻ

nên (2x-1;y)∈{(1;54);(3;18);(9;6);(27;2)}

=>(2x;y)∈{(2;54);(4;18);(10;6);(28;2)}

=>(x;y)∈{(1;54);(2;18);(5;6);(14;2)}

b: \(\frac{x}{6}-\frac{2}{y}=\frac{1}{30}\)

=>\(\frac{x}{6}-\frac{1}{30}=\frac{2}{y}\)

=>\(\frac{5x-1}{30}=\frac{2}{y}\)

=>y(5x-1)=60

mà 5x-1 chia 5 dư 4

nên (5x-1;y)∈{(4;15)}

=>(5x;y)∈{(5;15)}

=>(x;y)∈{(1;15)}

a: \(\frac{x}{2}-\frac{4}{y}=\frac15\)

=>\(\frac{x}{2}-\frac15=\frac{4}{y}\)

=>\(\frac{5x-2}{10}=\frac{4}{y}\)

=>y(5x-2)=40

mà 5x-2 chia 5 dư 3

nên (5x-2;y)∈{(8;5)}

=>(5x;y)∈{(10;5)}

=>(x;y)∈{(2;5)}

a. 3x² - 2x - 1 = 0

<=> 3x² - 3x + x - 1 = 0

<=> 3x(x - 1) + (x - 1) = 0

<=> (3x + 1)(x - 1) = 0

<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)

b. \(\dfrac{x+1}{3}+\dfrac{2x+3}{5}=\dfrac{3}{4}\)

<=> \(\dfrac{20\left(x+1\right)}{60}+\dfrac{12\left(2x+3\right)}{60}=\dfrac{45}{60}\)

<=> 20x + 20 + 24x + 36 = 45

<=> 44x = -11

<=> x = \(-\dfrac{1}{4}\)

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

    \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

b) Pt\(\Rightarrow\)\(5\cdot4\left(x+1\right)+3\cdot4\cdot\left(2x+3\right)=3\cdot3\cdot5\)

       \(\Leftrightarrow44x=-11\Rightarrow x=-\dfrac{1}{4}\)

a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)

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

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

Sao lm mỗi câu a, c thế

Câu c viết dấu hoặc đi