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

Ta có: \(x^4+4=5x^2\)

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

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

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

a) \(x^2-x+x=4\)

\(x^2=4\)

\(x=\pm2\)

b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)

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

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

c) Ta có: \(a+b+c=5-3-2=0\)

\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)

d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :

\(t^2-11t+18=0\)

\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)

\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)

\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)

a) Ta có: B(x)-M(x)=A(x)

nên M(x)=B(x)-A(x)

\(=x^4-2x^3+5x^2+x+10-x^4-2x^3+5x^2+3x+6\)

\(=-4x^3+10x^2+4x+16\)

a:Ta có: \(x\left(x-1\right)+x=4\)

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

\(\Leftrightarrow x^2=4\)

hay \(x\in\left\{2;-2\right\}\)

b: Ta có: \(3x\left(x-5\right)-2x+10=0\)

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

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

c: Ta có: \(5x^2-3x-2=0\)

\(\Leftrightarrow5x^2-5x+2x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d: Ta có: \(x^4-11x^2+18=0\)

\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)

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

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

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

a) x(x-1)+x=4

⇔x²=4⇔\(x=\pm2\)

b)3x(x-5)-2x+10=0

⇔3x(x-5)-2(x-5)=0

⇔(x-5)(3x-1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

c)5x²-3x-2=0

⇔ 5x(x-1)+2(x-1)=0

⇔ (x-1)(5x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)

d)x⁴-11x²+18=0

⇔ x²(x²-2)-9(x²-2)=0

⇔ (x²-2)(x²-9)=0

\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)

Đáp án: D

(x² - 4) (x² - 1) = 0 ⇔ x = ±2; x =  ±1 nên A = {-2; -1; 1; 2}

(x² - 4) (x² + 1) = 0 ⇔ x² - 4 = 0 ⇔ x = ±2 nên B = {-2;  2}

x⁴ - 5x² + 4)/x = 0 ⇔ x⁴ - 5x² + 4 = 0 ⇔ x = ±2; x =  ±1 nên D = {-2; -1; 1; 2}

=> A = D

x4 – 5x2 + 4 = 0 (1)

Đặt x2 = t, điều kiện t ≥ 0.

Khi đó (1) trở thành : t2 – 5t + 4 = 0 (2)

Giải (2) : Có a = 1 ; b = -5 ; c = 4 ⇒ a + b + c = 0

⇒ Phương trình có hai nghiệm t1 = 1; t2 = c/a = 4

Cả hai giá trị đều thỏa mãn điều kiện.

+ Với t = 1 ⇒ x2 = 1 ⇒ x = 1 hoặc x = -1;

+ Với t = 4 ⇒ x2 = 4 ⇒ x = 2 hoặc x = -2.

Vậy phương trình (1) có tập nghiệm S = {-2 ; -1 ; 1 ; 2}.

Bài 1:

\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)

\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)

\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)

Bài 2:

\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)

\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)

Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:

\(\left(y-1\right)\left(y+1\right)=120\)

\(\Leftrightarrow y^2-1=120\)

\(\Leftrightarrow y^2=121\)

\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)

+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)

\(\Leftrightarrow x^2+5x-6=0\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)

+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)

\(\Leftrightarrow x^2+5x+16=0\)

\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)

\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)

Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)

\(\Rightarrow\) loại

Vậy \(x\in\left\{1;-6\right\}\).

\(b,\) Đề thiếu vế phải rồi bạn.

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

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

\(=\left(x+y\right)\left(5x-1\right)\)

c:\(-6x^2+7x-2\)

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

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

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

1.

a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)

\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

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

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

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

b) Đặt t = x² ( t ≥ 0) ta có pt:

t² - t² - 2= 0

Δ= (-1)² - 4.1. (-2)

  = 9 > 0

⇒ \(\sqrt{\Delta}=\sqrt{9}=3\)

Vậy pt có 2 n_o phân biệt

x₁= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-\left(-1\right)+3}{2.1}=2\)

x₂= \(\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-\left(-1\right)-3}{2.1}=-1\)

Với t = 2 thì x²= 2 ⇔ x_1;2 = \(\pm4\)

Với t = -1 thì x²= -1 ⇔ x_3;4 ∈ ∅

Vậy tập nghiệm của pt là: S= \(\left\{\pm4\right\}\)

c) Đặt t = x² ( t ≥ 0) ta có pt:

4t² - 5t² - 9= 0

Δ= (-5)² - 4.4. (-9)

  = 169 > 0

⇒ \(\sqrt{\Delta}\) = \(\sqrt{169}=13\)

Vậy pt có 2 n_o phân biệt

x₁= \(\dfrac{5+13}{2.4}=\dfrac{9}{4}\)

x₂= \(\dfrac{5-13}{2.4}=-1\)

Với t = \(\dfrac{9}{4}\)  thì x²= \(\dfrac{9}{4}\) ⇔ x_1;2 = \(\pm\dfrac{3}{2}\)

Với t = -1 thì x²= -1 ⇔ x_3;4 ∈ ∅

Vậy tập nghiệm của pt là: S= \(\left\{\pm\dfrac{3}{2}\right\}\)

a: =>\(\dfrac{x+1-2x}{x\left(x+1\right)}=1\)

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

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

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

=>\(x=-1\pm\sqrt{2}\)

b: =>x^4+2x^2-x^2-2=0

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

=>x^2-1=0

=>x^2=1

=>x=1 hoặc x=-1

c: =>4x^4-9x^2+4x^2-9=0

=>(4x^2-9)(x^2+1)=0

=>4x^2-9=0

=>x=3/2 hoặc x=-3/2