sao đề nhìn bá vậy bạn ...

bài này chắc đặt \(\sqrt{x^3-3x+6}\)cho nó gọn thôi

a: =>x^2+4x-4x+1=0

=>x^2+1=0

=>Loại

b: =>2x-6+4=2x+2

=>-2=2(loại)

c: =>2(x+3)-2x-1=1

=>6-1=1

=>5=1(loại)

d =>x+3=0

=>x=-3(loại)

e: =>x^2-3x^2+3x-3x-2=0

=>-2x^2-2=0

=>x^2+1=0

=>Loại

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

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

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

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

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

`12)x(x-3)+3(x-3)=0`

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

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

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

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

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

a) Ta có: \(2x^3+5x^2-3x=0\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

b) Ta có: \(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)

\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)

c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)

\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)

\(\Leftrightarrow12x^2+15x-18=0\)

\(\Leftrightarrow12x^2+24x-9x-18=0\)

\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)

Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!

đây không phải Toán Lớp lớp 1 đâu

đâu phải toán lớp 1

bạn chọn nhầm à

Rảnh rỗi thật sự .-.

a: =(x-3)(2x+5)

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

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

x³ + 3x² + 3x = 7

<=> x³ + 3x² + 3x - 7 = 0

<=> (x - 1)(x² + 4x + 7) = 0

<=> x - 1 = 0 hoặc x² + 4x + 7 khác 0

<=> x - 1 = 0

<=> x = 1

a) ( x² + 3 x + 2 ) . ( x² + 3x+ 3 ) - 2 =0

<=>x⁴ + 3x³ + 3x² + 3x³ + 9x² + 9x + 2x² + 6x + 6 - 2 = 0

<=> x⁴ +  6x³  + 14x² + 15x + 4 = 0

<=> x⁴ + 3x³ + 3x³ + x² + 9x² + 4x² + 3x + 12x + 4 = 0

<=> x² . ( x² +3x + 1 ) + 3x . ( x² +3x + 1 ) + 4. ( x² + 3x + 1 ) = 0

<=> ( x² + 3x + 1 ) . ( x² + 3x + 4 ) = 0

<=> \(\orbr{\begin{cases}x^2+3x+1=0\\x^2+3x+4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

          \(x\notinℝ\)

<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

Nghiệm cuối cùng là : x₁ = \(\frac{-3+\sqrt{5}}{2}\);x₂ = \(\frac{-3-\sqrt{5}}{2}\)

b) ( x + 1 ) . ( x + 2 ) . ( x + 3 ) . ( x + 4 )  - 24 = 0

<=> ( x² + 2x + x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0

<=> ( x² + 3.x + 2 ) . ( x+3) . ( x + 4 ) -24         = 0

<=> ( x³ + 3.x ² + 3.x² + 9x + 2x + 6 ) . ( x + 4 ) - 24 = 0 

<=> ( x³ + 3x + 2 ) . ( x + 3 ) .( x + 4 ) = 0

<=> ( x³ + 3x² + 3x² + 9x + 2x + 6 ) . ( x + 4) - 24 = 0

<=> ( x³ + 6.x² + 11.x + 6 ) . ( x + 4 ) -24 = 0 

<=> x⁴ + 4.x³ + 6.x³ + 24.x² + 11.x² + 44.x + 6.x + 24 - 24 =0

<=> x⁴ + 10.x³+ 35. x²  + 50.x = 0

<=> x. ( x³ + 10.x² + 35 .x + 50 ) = 0

<=> x. ( x³ + 5.x² +5.x² + 25.x+ 10 + 50 ) = 0

<=> x. [ x² . ( x+5 ) + 5.x. ( x+5 ) + 10.( x + 5 ) ] = 0

<=> x. ( x + 5 ) . ( x² + 5.x + 10 ) = 0

=> \(\hept{\begin{cases}x=0\\x+5=0\\x^2+5.x+10=0\end{cases}}\)

=> \(\hept{\begin{cases}x=0\\x=-5\\x\notinℝ\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)

Nghiệm cuối cùng là : x₁ = -5 ; x₂ = 0

c) x³ + 3.x² + 3x = 7

<=> x³ + 3.x² + 3x - 7 = 0

<=> ( x + 1 )³ - 8          = 0

<=> ( x + 1 )³               = 8

<=> ( x + 1 ) ³               = 2³ 

<=> x + 1                      = 2

<=> x                              =1

Vậy x = 1