haizz mà đứa trong hình là con nhà ai mà dễ thương wa

pt quá vĩ đại =.= cx trên OLM lun 

a)x⁵+x-1=0

<=>(x⁵+x⁴+x³+x²+x)-(x⁴+x³+x²+x+1)=0

<=>(x⁴+x³+x²+x+1)(x-1)=0

Do x⁴+x³+x²+x+1>0

=>x+1=0

<=>x=1

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

cái này bạn cố gắng phân tích ra đi

6x⁴ - x³ - 7x² + x + 1 = 0

=> (x + 1)(3x + 1)(x - 1)(2x - 1) = 0

=> x + 1 = 0 => x = -1

hoặc 3x + 1 = 0 => x = -1/3

hoặc x - 1 = 0 => x = 1

hoặc 2x - 1 = 0 => x = 1/2

Vậy x = -1, x = -1/3, x = 1 , x = 1/2

a. Ta có:

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-3=\sqrt{6}\\x-3=-\sqrt{6}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{6}\\x=3-\sqrt{6}\end{matrix}\right.\)

Ta có:

\(x^2-7x+14=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{7}{2}+\dfrac{49}{4}+\dfrac{7}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{7}{2}\right)^2+\dfrac{7}{4}=0\)

Ta có: \(\left(x+\dfrac{7}{2}\right)^2\ge0\)

=> \(\left(x+\dfrac{7}{2}\right)^2+\dfrac{7}{4}>0\)

=> pt vô nghiệm

Lạnh xun loz

phải không mày >

haizz

\(6x^4+7x^3-36x^2-7x+6=0\)

\(\Leftrightarrow\left(6x^4-11x^3-3x^2+2x\right)+\left(18x^3-33x^2-9x+6\right)=0\)

\(\Leftrightarrow x\left(6x^3-11x^2-3x+2\right)+3\left(6x^3-11x^2-3x+2\right)=0\)

\(\Leftrightarrow\left(6x^3-11x^2-3x+2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[\left(6x^3-14x+4x\right)+\left(3x^2-7x+2\right)\right]\left(x+3\right)=0\)

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

\(\Leftrightarrow\left(3x^2-7x+2\right)\left(2x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(3x^2-6x-x+2\right)\left(2x+1\right)\left(x+3\right)=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x-1=0\end{cases}}\)hoặc \(\orbr{\begin{cases}2x+1=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}\)hoặc\(\orbr{\begin{cases}x=\frac{-1}{2}\\x=-3\end{cases}}\)

Vậy tập hợp nghiệm \(S=\left\{2;-3;\frac{1}{3};\frac{-1}{2}\right\}\)

a: ĐKXĐ: \(x^2-6x+6\ge0\)

=>\(x^2-6x+9-3\ge0\)

=>\(\left(x-3\right)^2-3\ge0\)

=>\(\left(x-3\right)^2\ge3\)

=>\(\left[\begin{array}{l}x-3\ge\sqrt3\\ x-3\le-\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge\sqrt3+3\\ x\le-\sqrt3+3\end{array}\right.\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

=>\(x^2-6x+6-4\cdot\sqrt{x^2-6x+6}+3=0\)

=>\(\left(\sqrt{x^2-6x+6}-3\right)\left(\sqrt{x^2-6x+6}-1\right)=0\)

TH1: \(\sqrt{x^2-6x+6}-3=0\)

=>\(\sqrt{x^2-6x+6}=3\)

=>\(x^2-6x+6=9\)

=>\(x^2-6x-3=0\)

=>\(x^2-6x+9-12=0\)

=>\(\left(x-3\right)^2=12\)

=>\(\left[\begin{array}{l}x-3=2\sqrt3\\ x-3=-2\sqrt3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\sqrt3+3\left(nhận\right)\\ x=3-2\sqrt3\left(nhận\right)\end{array}\right.\)

TH2: \(\sqrt{x^2-6x+6}-1=0\)

=>\(x^2-6x+6=1\)

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

=>(x-1)(x-5)=0

=>\(\left[\begin{array}{l}x=1\left(nhận\right)\\ x=5\left(nhận\right)\end{array}\right.\)

b: ĐKXĐ: x∈R

\(x^2-x+8-4\sqrt{x^2-x+4}=0\)

=>\(x^2-x+4-4\cdot\sqrt{x^2-x+4}+4=0\)

=>\(\left(\sqrt{x^2-x+4}-2\right)^2=0\)

=>\(\sqrt{x^2-x+4}-2=0\)

=>\(\sqrt{x^2-x+4}=2\)

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

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

=>x(x-1)=0

=>x=0 hoặc x=1

c: \(x^2+\sqrt{4x^2-12x+44}=3x+4\)

=>\(x^2-3x-4+2\sqrt{x^2-3x+11}=0\)

=>\(x^2-3x+11+2\sqrt{x^2-3x+11}-15=0\)

=>\(\left(\sqrt{x^2-3x+11}+5\right)\left(\sqrt{x^2-3x+11}-3\right)=0\)

=>\(\sqrt{x^2-3x+11}-3=0\)

=>\(\sqrt{x^2-3x+11}=3\)

=>\(x^2-3x+11=9\)

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

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

=>x=1(nhận) hoặc x=2(nhận)

g: =>(x-1)(x-2)=0

=>x=1 hoặc x=2

i: \(\Leftrightarrow x^4-x^3+x^3-x^2+2x^2-2x+8x-8=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

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

=>x=1 hoặc x=-2

câu i vs câu h nữa