Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

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Bài `2:`

`b)3x-15=2x(x-5)`

`<=>3(x-5)-2x(x-5)=0`

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

`<=>(2x-5-x-2)(2x-5+x+2)=0`

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

`<=>(2x-1-2)(2x-1+2)=0`

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

\(\frac{{x + 5}}{3} = 1 - \frac{{x - 2}}{4}\)

\(\frac{{\left( {x + 5} \right).4}}{{3.4}} = \frac{{12}}{{12}} - \frac{{\left( {x - 2} \right).3}}{{4.3}}\)

\(\frac{{4x + 20}}{{12}} = \frac{{12}}{{12}} - \frac{{3x - 6}}{{12}}\)

\(4x + 20 = 12 - \left( {3x - 6} \right)\)

\(4x + 20 = 12 - 3x + 6\)

\(4x + 3x = 12 + 6 - 20\)

\(7x =  - 2\)

\(x = \left( { - 2} \right):7\)

\(x = \frac{{ - 2}}{7}\)

Vậy phương trình có nghiệm là \(x = \frac{{ - 2}}{7}\).

Trừ vế cho vế:

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

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

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

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

Thế vào pt đầu:

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=0\\x^2+5x+29=0\left(vô-nghiệm\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=4\Rightarrow y=4\end{matrix}\right.\)

X=-2,3,1/3

\(6x^4-5x^3-38x^2-5x+6=0\)

\(\Leftrightarrow6x^4-12x^3+17x^3-34^2-4x^2+8x-3x+6=0\)

\(\Leftrightarrow6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-4x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x^3+18x^2-x^2-3x-x-3=0\right)\)

\(\Leftrightarrow\left(x-2\right)\left[6x^2\left(x+3\right)-x\left(x+3\right)-\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(6x^2-3x+2x-1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left[6x\left(x-\frac{1}{2}\right)+2\left(x-\frac{1}{2}\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x-\frac{1}{2}\right)\left(6x+2\right)=0\)

kết quả là bao nhiêu zậy Lê Mạnh Tiến Đạt

b) (0,5 - x)² - 3 = 0 ⇔ 0,5² + x² - 2.0,5x - 3 = 0

                               ⇔ x² + 0,25 - x - 3 = 0

                               ⇔ x² - x - 2,75 = 0

                               ⇔ ????

c: \(\Leftrightarrow\left(2x-3\sqrt{2}\right)\left(2x+\sqrt{2}\right)=0\)

hay \(x\in\left\{\dfrac{3\sqrt{2}}{2};-\dfrac{\sqrt{2}}{2}\right\}\)