Question

[Gia Huỳnh]

Giúp với em với ạ đang cần gấp

Answer 1

[Yen Nhi]

`Answer:`

a. \(x^3+x^2-x+2=4x-1\)

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

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

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)

b. \(x^4+35x^2-74=0\)

\(\Leftrightarrow x^4+37x^2-2x^2-74=0\)

\(\Leftrightarrow x^2.\left(x^2+37\right)-2\left(x^2+37\right)=0\)

\(\Leftrightarrow\left(x^2-2\right)\left(x^2+37\right)=0\)

Mà \(x^2+37\ne0\forall x\)

\(\Rightarrow x^2-2=0\)

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow x=\pm\sqrt{2}\)

c. \(2x^4+2x^3-76x^2+4x+8=0\)

\(\Leftrightarrow2\left(x^4+x^3-38x^2+2x+4\right)=0\)

\(\Leftrightarrow x^4+x^3-38x^2+2x+4=0\)

\(\Leftrightarrow x^4+7x^3+2x^2-6x^3-42x^2-12x+2x^2+14x+4=0\)

\(\Leftrightarrow x^2.\left(x^2+7x+2\right)-6x\left(x^2+7x+2\right)+2.\left(x^2+7x+2\right)=0\)

\(\Leftrightarrow\left(x^2-6x+2\right)\left(x^2+7x+2\right)=0\)

Trường hợp 1: \(x^2-6x+2\)

Ta có \(\Delta'=\left(-3\right)^2-2=7>0\)

Vậy phương trình có hai nghiệm phân biệt:

\(x_1=3-\sqrt{7}\)

\(x_2=3+\sqrt{7}\)

Trường hợp 2: \(x^2+7x+2=0\)

Ta có \(\Delta=7^2-4.2=41>0\)

Vậy phương trình có hai nghiệm phân biệt:

\(x_1=\frac{-7-\sqrt{41}}{2}\)

\(x_2=\frac{-7+\sqrt{41}}{2}\)

d. \(\left(3x+3\right)^4+\left(3x+5\right)^4=0\)

Mà \(\left(3x+3\right)^4\ge0;\left(3x+5\right)^4\ge0\forall x\inℝ\)

`=>` Để phương trình có nghiệm thì \(\hept{\begin{cases}3x+3=0\\3x+5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=-3\\3x=-5\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\x=-\frac{5}{3}\end{cases}}\text{(Vô lý)}\)

Vậy phương trình vô nghiệm.

e. \(\left(2x+1\right)\left(2x+3\right)\left(2x+5\right)\left(2x+7\right)=169\)

\(\Leftrightarrow[\left(2x+1\right)\left(2x+7\right)][\left(2x+3\right)\left(2x+5\right)]-169=0\)

\(\Leftrightarrow\left(4x^2+16x+6+1\right)\left(4x^2+16x+6-1\right)-169=0\)

Đặt \(a=4x^2+16x+6\)

\(\Rightarrow\left(a+1\right)\left(a-1\right)-169=0\)

\(\Leftrightarrow a^2-1^2-169=0\)

\(\Leftrightarrow a^2-170=0\)

\(\Leftrightarrow a=\pm\sqrt{170}\)

\(\Rightarrow4x^2+16+6=\pm\sqrt{170}\)

\(\Leftrightarrow4x^2+16+16-10=\pm\sqrt{170}\)

\(\Leftrightarrow4\left(x+2\right)^2=\pm\sqrt{170}+10\)

\(\Leftrightarrow\left(x+2\right)^2=\left(\pm\sqrt{\frac{\pm\sqrt{170}+10}{4}}\right)^2\)

\(\Leftrightarrow x+2=\pm\sqrt{\frac{\pm\sqrt{170}+10}{4}}\)

\(\Leftrightarrow x=\pm\sqrt{\frac{\pm\sqrt{170}+10}{4}}-2\)