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.

a: \(8x\left(x-2017\right)-2x+4034=0\)

\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

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) Tìm được x = -4.        

b) Tìm được x = 3.

c) Tìm được x = ±1.

a) x = 8 3 .                            b) x = − 9 20 .  

a) x = -1.                      b) x = 4 hoặc x = 5.

c) x = ± 2 .                  d) x = 1 hoặc x = 2.

a. \(\dfrac{3}{4}\) + \(\dfrac{-4}{5}\) - \(\dfrac{1}{2}\) = \(\dfrac{-1}{20}\)  - \(\dfrac{1}{2}\) = \(\dfrac{-11}{20}\) 

b. (4 - \(\dfrac{5}{12}\) ): 2 + \(\dfrac{5}{24}\) 

= \(\dfrac{43}{12}\) : 2 + \(\dfrac{5}{24}\) 

= \(\dfrac{43}{24}\) + \(\dfrac{5}{24}\) 

=\(\dfrac{48}{24}\) = 2

2.2 

\(\dfrac{4}{7}\) .x - \(\dfrac{2}{3}\) = \(\dfrac{1}{5}\) 

  \(\dfrac{4}{7}\) x   = \(\dfrac{1}{5}\) + \(\dfrac{2}{3}\) 

  \(\dfrac{4}{7}\) . x = \(\dfrac{13}{15}\) 

        x = \(\dfrac{91}{60}\)

2.1

\(a)\dfrac{3}{4}+\dfrac{-4}{5}-\dfrac{1}{2}\\ =\dfrac{3\times5}{4\times5}+\dfrac{-4\times4}{5\times4}-\dfrac{1\times10}{2\times10}\\ =\dfrac{15}{20}+\dfrac{-16}{20}-\dfrac{10}{20}\\ =\dfrac{15-16-10}{20}\\ =\dfrac{-11}{20}\)

\(b)\left(4-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{4\times12}{1\times12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48}{12}-\dfrac{5}{12}\right):2+\dfrac{5}{24}\\ =\left(\dfrac{48-5}{12}\right):2+\dfrac{5}{24}\\ =\dfrac{43}{12}:2+\dfrac{5}{24}\\ =\dfrac{43}{12}\times\dfrac{1}{2}+\dfrac{5}{24}\\ =\dfrac{43}{24}+\dfrac{5}{24}\\ =\dfrac{43+5}{24}\\ =\dfrac{48}{24}\\ =2\)

\(2.2\\ \dfrac{4}{7}\times x-\dfrac{2}{3}=\dfrac{1}{5}\\ \dfrac{4}{7}\times x=\dfrac{1}{5}+\dfrac{2}{3}\\ \dfrac{4}{7}\times x=\dfrac{1\times3}{5\times3}+\dfrac{2\times5}{3\times5}\\ \dfrac{4}{7}\times x=\dfrac{3}{15}+\dfrac{10}{15}\\ \dfrac{4}{7}\times x=\dfrac{13}{15}\\ x=\dfrac{13}{15}:\dfrac{4}{7}\\ x=\dfrac{13}{15}\times\dfrac{7}{4}\\ x=\dfrac{91}{60}\)

a) Kết quả M = x 4 – 1.

b) Kết quả M =  x 2  – 2x – 3.

\(\dfrac{4}{5}-\left|x+\dfrac{1}{2}\right|=\dfrac{1}{4}\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{11}{20}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{11}{20}\\x+\dfrac{1}{2}=-\dfrac{11}{20}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{11}{20}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{11}{20}\left(x\ge-\dfrac{1}{2}\right)\\x+\dfrac{1}{2}=-\dfrac{11}{20}\left(x< -\dfrac{1}{2}\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\left(tm\right)\\x=-\dfrac{21}{20}\left(tm\right)\end{matrix}\right.\)

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