PT <=> \(x^4+4x^3+6x^2+4x+1=0\)

Bạn giải rõ ràng ra đc ko ?

x(x+2)(x²+2x+2)+1=0

<=>(x²+2x)(x²+2x+2)+1=0

Đặt x²+2x=a

PT <=>a(a+2)+1=0

      <=>a²+2a+1=0

       <=> (a+1)²=0

       <=>a= -1

=>x²+2x= -1

<=>x²+2x+1=0

<=>( x+1)²=0

<=>x= -1

\(x\left(x+2\right)\left(x^2+2x+2\right)+1=0\Leftrightarrow\left(x+1-1\right)\left(x+1+1\right)\left(x^2+2x+1+1\right)+1=0\) \(Đạt:x+1=a\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)+1=0\Leftrightarrow\left(a^2-1\right)\left(a^2+1\right)+1=0\Leftrightarrow a^4-1+1=0\Leftrightarrow a^4=0\Leftrightarrow a=0\Leftrightarrow x=-1.Vậy:x=-1\)

\(\Leftrightarrow x^2+4=2x+3\)

=>x^2-2x+1=0

=>(x-1)^2=0

=>x=1

Đặt \(\sqrt{6x^2-12x+7}=t\left(t\ge0\right)\)

<=>\(t^2-7=6x^2-12x\)

\(\Leftrightarrow\dfrac{t^2-7}{6}=x^2-2x\)

Ta có pt mới:

\(\dfrac{7-t^2}{6}+t=0\)

\(\Leftrightarrow t^2-6t-7=0\)

\(\Leftrightarrow t^2-2\cdot t\cdot3+9-9-7=0\)

\(\Leftrightarrow\left(t-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}t=7\\t=-1\end{matrix}\right.\)(loại t=-1)

Với t=7

=>\(\sqrt{6x^2-12x+7}=7\)

<=>6x²-12x+7=49

<=>6x²-12x-42=0

<=>x²-2x-7=0

<=>(x-1)²=8

=>\(\left[{}\begin{matrix}x=1+2\sqrt{2}\\x=1-2\sqrt{2}\end{matrix}\right.\)

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

3x-15=2x²-10x

10x-3x=15+2x²

7x=15+2x²

7x-2x*x=15

5x*x=15

x*x=15/5

x*x=3

=> x\(\in\)rỗng(ký hiệu)

Mình cũng không chắc nữa do ms học lớp 6 thôi

Chúc bạn học tốt!^_^

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

<=> 3x -15 =2x²-10x

<=>2x²-13x+15=0

<=>x=5, x= 3/2

\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\) (ĐKXĐ: \(x\ne0;x\ne\dfrac{1}{2}\))

\(\)\(\Leftrightarrow\dfrac{x+3}{x}=\dfrac{2\left(x+1\right)}{2x-1}\Leftrightarrow\left(x+3\right)\left(2x-1\right)=2x\left(x+1\right)\)

\(\Leftrightarrow2x^2+6x-x-3=2x^2+2x\)

\(\Leftrightarrow2x^2-2x^2+6x-x-2x=3\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\left(TM\right)\)

\(\Rightarrow S=\left\{1\right\}\)

\(\dfrac{x+3}{x}=\dfrac{2x+2}{2x-1}\)

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

\(\Leftrightarrow2x^2-x+6x-3=2x^2+2x\)

\(\Leftrightarrow2x^2+5x-3-2x^2-2x=0\)

\(\Leftrightarrow3x-3=0\)

\(\Leftrightarrow3\left(x-1\right)=0\)

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy \(S=\left\{1\right\}\)

ĐKXĐ: x >= 2

Pt <=> 3sqrt(x - 2) = 6

<=> sqrt(x - 2) = 2

<=> x - 2 = 4

<=> x = 6 (thỏa ĐKXĐ)

`a,4x^2+(x-1)^2-(2x+1)^2=0`

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

`<=>x(4x-3x-6)=0`

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

`<=>` $\left[ \begin{array}{l}x=0\\x=6\end{array} \right.$

`b)(x^2-3x)^2+5(x^2-3x)+6=0`
Đặt `x^2-3x=a(a>=-9/4)`
`pt<=>a^2+5a+6=0`
`<=>(a+2)(a+3)=0`
`<=>` $\left[ \begin{array}{l}a=-2\\a=-3(l)\end{array} \right.$
`<=>x^2-3x=-2`
`<=>x^2-3x+2=0`
`<=>(x-1)(x-2)=0`
`<=>` $\left[ \begin{array}{l}x=2\\x=1\end{array} \right.$

Bạn cần lời giải hay cách giải

Bài làm:

Ta có: \(y^2+4^x+2y-2^{x+1}+2=0\)

\(\Leftrightarrow\left(y^2+2y+1\right)+\left(2^{2x}-2^{x+1}+1\right)=0\)

\(\Leftrightarrow\left(y+1\right)^2+\left[\left(2^x\right)^2-2.2^x+1\right]=0\)

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

Mà \(\hept{\begin{cases}\left(y+1\right)^2\ge0\\\left(2^x-1\right)^2\ge0\end{cases}}\forall x,y\)

\(\Rightarrow\left(y+1\right)^2+\left(2^x-1\right)^2\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(y+1\right)^2=0\\\left(2^x-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=-1\\2^x=1=2^0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\y=-1\end{cases}}\)

Vậy \(\left(x;y\right)=\left(0;-1\right)\)

Cảm ơn bạn nhiều nha !