1.

a/ \(\Leftrightarrow\left(x+1\right)\left(x^2+3x+2\right)+\left(x-1\right)\left(x^2-3x+2\right)-12=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+2\right)+3x\left(x+1\right)-3x\left(x-1\right)+\left(x-1\right)\left(x^2+2\right)-12=0\)

\(\Leftrightarrow2x\left(x^2+2\right)+6x^2-12=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x^2+4x+6\right)=0\Rightarrow x=1\)

b/ Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)

\(x^2+\frac{1}{x^2}+3\left(x+\frac{1}{x}\right)+4=0\)

Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)

\(t^2-2+3t+4=0\Rightarrow t^2+3t+2=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}=-1\\x+\frac{1}{x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2+x+1=0\left(vn\right)\\x^2+2x+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)

1c/

\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)

\(\Leftrightarrow x^4\left(x+1\right)-2x^3\left(x+1\right)+5x^2\left(x+1\right)-2x\left(x+1\right)+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x^4-2x^3+5x^2-2x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x^4-2x^3+x^2+x^2-2x+1+3x^2=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x=0\\x-1=0\\x=0\end{matrix}\right.\) \(\Rightarrow\) không tồn tại x thỏa mãn

Vậy pt có nghiệm duy nhất \(x=-1\)

2.

a. \(x^4-x^3+x^2+x^2-x+1=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(vn\right)\\x^2-x+1=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\left(vn\right)\end{matrix}\right.\)

Vậy pt vô nghiệm

b.

\(x^4+x^3+x^2+x+1=0\)

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

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

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

Mà \(\left\{{}\begin{matrix}\left(x+1\right)^2\left(x^2-x+1\right)\ge0\\x^2\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi: \(\left\{{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\) ko tồn tại x thỏa mãn

Chọn B

+, Nếu x = 0 => ko tồn tại y thuộc Z

+, Nếu x khác 0 => x^2 >= 1 => x^2-1 >= 0

Có : y^3 = x^3+2x^2+3x+2 > x^3 ( vì 2x^2+3x+2 > 0 )

Lại có : y^3 = (x^3+3x^3+3x+1)-(x^2-1) = (x+1)^3 - (x^2-1) < = (x+1)^3

=> x^3 < y^3 < = (x+1)^3

=> y^3 = (x+1)^3

=> x^2-1 = 0

=> x=-1 hoặc x=1

+, Với x=-1 thì y = 0

+, Với x=1 thì y = 2

Vậy .............

Tk mk nha

Ta có: \(x^3+2x^2+3x+2=y^3\)                             (1)

Xét \(2x^2+3x+2=2\left(x^2+\frac{3}{2}x\right)+2=2\left(x^2+\frac{3}{2}x+\frac{9}{16}\right)+2-2.\frac{9}{16}\)

\(=2\left(x+\frac{3}{4}\right)^2+\frac{7}{8}\) Vì \(\left(x+\frac{3}{4}\right)^2\ge0\Rightarrow2\left(x+\frac{3}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}>0\)

\(\Rightarrow y^3>x^3\Rightarrow y^3\ge\left(x+1\right)^3\)

\(\Rightarrow x^3+2x^2+3x+2\ge\left(x+1\right)^3\) \(\Rightarrow x^3+2x^2+3x+2\ge x^3+3x^2+3x+1\)

\(\Rightarrow x^3+3x^2+3x+1-x^3-2x^2-3x-2\le0\)

\(\Rightarrow x^2-1\le0\Rightarrow x^2\le1\) Vì \(x\in Z\Rightarrow\orbr{\begin{cases}x^2=1\\x^2=0\end{cases}}\)

+ TH1: x² = 0 => x =0 Thay vào pt (1) ta được y³ = 2 (loại) vì y nguyên

+ TH2 : x² = 1 => \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Thay x=1 vào pt (1) ta đc: 1+2+3+2 = 8 = y³ => y = 2

Thay x= -1 vào pt (1) ta đc: -1 + 2 -3 +2 = 0 =y³ => y = 0

Vậy cặp (x;y) là (1;2) ; (-1;0).

\(Xét \(2x^2+3x+2=2\left(x+\dfrac{3}{4}\right)^2+\dfrac{7}{16}>0\forall x\in R\) => \(x^3< y^3\left(1\right)\) (1) Giả sử : \(y^3< \left(x+2\right)^3\) \(\Leftrightarrow x^3+2x^2+3x+2< x^3+6x^2+12x+8\) \(\Leftrightarrow-4x^2-9x-6< 0\) \(\Leftrightarrow4x^2+9x+6>0\) \(\Leftrightarrow4\left(x+\dfrac{9}{8}\right)^2+\dfrac{15}{64}>0\) => Giả sử đúng . => \(y^3< \left(x+2\right)^3\left(2\right)\) Từ (1)(2) => \(y^3=\left(x+1\right)^3\) \(\Leftrightarrow x^3+2x^2+3x+2=x^3+3x^2+3x+1\) \(\Leftrightarrow x^2=1\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\) .) Khi \(x=1\Rightarrow y=2\). .) Khi \(x=-1\Rightarrow y=0\) Vậy nghiệm của pt ( x;y ) = {( 1;2 ) ; ( -1;0 )}\)

a, Sửa đề:

\(3x^2-\sqrt3 x+\dfrac14(dkxd:x\geq0)\\=(x\sqrt3)^2-2\cdot x\sqrt3\cdot\dfrac12+\Bigg(\dfrac12\Bigg)^2\\=\Bigg(x\sqrt3-\dfrac12\Bigg)^2\)

b, 

\(x^2-x-y^2+y\\=(x^2-y^2)-(x-y)\\=(x-y)(x+y)-(x-y)\\=(x-y)(x+y-1)\)

c,

\(x^4+x^3+2x^2+x+1\\=(x^4+x^3+x^2)+(x^2+x+1)\\=x^2(x^2+x+1)+(x^2+x+1)\\=(x^2+x+1)(x^2+1)\)

d,

\(x^3+2x^2+x-16xy^2\\=x(x^2+2x+1-16y^2)\\=x[(x+1)^2-(4y)^2]\\=x(x+1-4y)(x+1+4y)\\Toru\)

Đặt x2−2x+m=tx2−2x+m=t, phương trình trở thành t2−2t+m=xt2−2t+m=x

Ta có hệ {x2−2x+m=tt2−2t+m=x{x2−2x+m=tt2−2t+m=x

⇒(x−t)(x+t−1)=0⇒(x−t)(x+t−1)=0

⇔[x=tx=1−t⇔[x=tx=1−t

⇔[x=x2−2x+mx=1−x2+2x−m⇔[x=x2−2x+mx=1−x2+2x−m

⇔[m=−x2+3xm=−x2+x+1⇔[m=−x2+3xm=−x2+x+1

Phương trình hoành độ giao điểm của y=−x2+x+1y=−x2+x+1 và y=−x2+3xy=−x2+3x:

−x2+x+1=−x2+3x−x2+x+1=−x2+3x

⇔x=12⇒y=54⇔x=12⇒y=54

Đồ thị hàm số y=−x2+3xy=−x2+3x và y=−x2+x+1y=−x2+x+1: 

a) x = 0                 b) x = - 1 3

c) x = 28 15               d) x = -82.