câu 1

x=5

x=4

câu 2

x=2

x=-5

a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

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

Ta có: \(x^2-4x+5\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

hay \(x^2-4x+5>0\forall x\)(2)

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

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

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

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

Vậy: x∈{-3;5}

d) Ta có: \(\left(x^2-1\right)^2=4x+1\)

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

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

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

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

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

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

Ta có: \(x^2+2x+2\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy: x∈{0;2}

x²-5x-4x+20=0

x²-4x-5x+20=0

x(x-4)-5(x-4)=0

(x-4)(x-5)=0

\(\Rightarrow\hept{\begin{cases}x-4=0\Rightarrow x=4\\x-5=0\Rightarrow x=5\end{cases}}\)

tớ hỏi nốt nha

tính nhanh:20²-19²+18²-17²+...+2²-1²

\(x^2-9x+20=0\)

\(\Rightarrow x^2-2.x.\frac{9}{2}+\frac{81}{4}-\frac{1}{4}=0\)

\(\Rightarrow x^2-2.x.\frac{9}{2}+\left(\frac{9}{2}\right)^2-\frac{1}{4}=0\)

\(\Rightarrow\left(x-\frac{9}{2}\right)^2-\left(\frac{1}{2}\right)^2=0\)

\(\Rightarrow\left(x-\frac{9}{2}-\frac{1}{2}\right)\left(x-\frac{9}{2}+\frac{1}{2}\right)=0\)

\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy \(x=5\)hoặc   \(x=4\)

<=>x^2-4x-5x+20=0

<=>x(x-4)-5(x-4)=0

<=>(x-5)(x-4)=0

=>x=5; x=4

\(x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow\left(x^2-4x\right)-\left(5x-20\right)=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

Vậy \(x=4\) hoặc \(x=5\)

x²-9x+20=0

<=>x² - 4x-5x +20 =0

<=>x(x-4) -5(x-4)=0

<=>(x-5)(x-4) =0

<=> x-5=0 hoặc x-4=0

<=> x=5 hoặc x=4

Vậy x=5 hoặc x=4

Pt\(\Leftrightarrow\left(x^4-4x^2\right)-\left(5x^2-20\right)=0\Leftrightarrow\left(x^2-4\right)\left(x^2-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=4\\x^2=5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2;x=-2\\x=\sqrt{5};x=-\sqrt{5}\end{cases}}}\)

Vì x nguyên dương nên \(\Rightarrow\orbr{\begin{cases}x=2\\x=\sqrt{5}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=2\\x=\sqrt{5}\end{cases}}\)

a) \(x^2-4=0\)

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

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

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

b) \(x\left(x+5\right)=9x\)

\(\Rightarrow x^2+5x-9x=0\)

\(\Rightarrow x^2-4x=0\)

\(\Rightarrow x\left(x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

c) \(3x^3-48x=0\)

\(\Rightarrow3x\left(x^2-16\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-16=0\Rightarrow\left(x-4\right)\left(x+4\right)=0\end{matrix}\right.\)

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

d) \(x^4+x^2-20=0\)

\(\Rightarrow\left(x^2\right)^2+x^2-20=0\)

Đặt x² = a

\(\Rightarrow a^2+a-20=0\)

\(\Rightarrow a^2+5a-4a-20=0\)

\(\Rightarrow a\left(a+5\right)-4\left(a+5\right)=0\)

\(\Rightarrow\left(a-4\right)\left(a+5\right)=0\)

\(\Rightarrow\left(x^2-4\right)\left(x^2+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x^2+5=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x^2=4\Rightarrow x=\pm2\\x^2=-5\Rightarrow x\in\varnothing\end{matrix}\right.\)

d) x⁴ + x² - 20 = 0

\(\Rightarrow\) x⁴ + x² = 20

\(\Rightarrow\) x⁴ + x² = 2⁴ + 2²

\(\Rightarrow\) x = 2

a) x² - 4 = 0

\(\Rightarrow\) x² = 4

\(\Rightarrow\) x = 2 hoặc -2

ta có: x^2-9x+20=0

suy ra: x^2-4x-5x+20=0

           x(x-4)-5(x-4)

           (x-5)(x-4)

suy ra: x=5;x=4

nhớ click nha

a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4

=>2*căn(x+5)=4

=>căn (x+5)=2

=>x+5=4

=>x=-1

b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

=>2*căn x-1=16

=>x-1=64

=>x=65

c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)

TH₁: \(x\ge3\)

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

TH₂: \(2\le x< 3\)

\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)

TH₃: \(0\le x< 2\)

\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

TH₄: \(x< 0\)

\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)

Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)