a) (3x² - 7x – 10)[2x² + (1 - √5)x + √5 – 3] = 0

=> hoặc (3x² - 7x – 10) = 0 (1)

hoặc 2x² + (1 - √5)x + √5 – 3 = 0 (2)

Giải (1): phương trình a - b + c = 3 + 7 - 10 = 0

nên

x₁ = - 1, x₂ = =

Giải (2): phương trình có a + b + c = 2 + (1 - √5) + √5 - 3 = 0

nên

x₃ = 1, x₄ =

b) x³ + 3x²– 2x – 6 = 0 ⇔ x²(x + 3) – 2(x + 3) = 0 ⇔ (x + 3)(x² - 2) = 0

=> hoặc x + 3 = 0

hoặc x² - 2 = 0

Giải ra x₁ = -3, x₂ = -√2, x₃ = √2

c) (x² - 1)(0,6x + 1) = 0,6x² + x ⇔ (0,6x + 1)(x² – x – 1) = 0

=> hoặc 0,6x + 1 = 0 (1)

hoặc x² – x – 1 = 0 (2)

(1) ⇔ 0,6x + 1 = 0

⇔ x₂ = =

(2): ∆ = (-1)² – 4 . 1 . (-1) = 1 + 4 = 5, √∆ = √5

x₃ = , x₄ =

Vậy phương trình có ba nghiệm:

x₁ = , x₂ = , x₃ = ,

d) (x² + 2x – 5)² = ( x² – x + 5)² ⇔ (x² + 2x – 5)² - ( x² – x + 5)² = 0

⇔ (x² + 2x – 5 + x² – x + 5)( x² + 2x – 5 - x² + x - 5) = 0

⇔ (2x² + x)(3x – 10) = 0

⇔ x(2x + 1)(3x – 10) = 0

Hoặc x = 0, x = , x =

Vậy phương trình có 3 nghiệm:

x₁ = 0, x₂ = , x₃ =

1. \(\sqrt{x^2-4}-x^2+4=0\)( ĐK: \(\orbr{\begin{cases}x\ge2\\x\le-2\end{cases}}\))

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\left(tm\right)\\x=\pm\sqrt{5}\left(tm\right)\end{cases}}\)

Vậy pt có tập no \(S=\left\{2;-2;\sqrt{5};-\sqrt{5}\right\}\)

2. \(\sqrt{x^2-4x+5}+\sqrt{x^2-4x+8}+\sqrt{x^2-4x+9}=3+\sqrt{5}\)ĐK: \(\hept{\begin{cases}x^2-4x+5\ge0\\x^2-4x+8\ge0\\x^2-4x+9\ge0\end{cases}}\)

\(\Leftrightarrow\sqrt{x^2-4x+5}-1+\sqrt{x^2-4x+8}-2+\sqrt{x^2-4x+9}-\sqrt{5}=0\)

\(\Leftrightarrow\frac{x^2-4x+4}{\sqrt{x^2-4x+5}+1}+\frac{x^2-4x+4}{\sqrt{x^2-4x+8}+2}+\frac{x^2-4x+4}{\sqrt{x^2-4x+9}+\sqrt{5}}=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}\right)=0\)

Từ Đk đề bài \(\Rightarrow\frac{1}{\sqrt{x^2-4x+5}+1}+\frac{1}{\sqrt{x^2-4x+8}+2}+\frac{1}{\sqrt{x^2}-4x+9+\sqrt{5}}>0\)

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

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy pt có no x=2

374

a: \(5x^2-8x=0\)

=>x(5x-8)=0

=>\(\left[\begin{array}{l}x=0\\ 5x-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ 5x=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac85\end{array}\right.\)

b: \(-3x^2-6x=0\)

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

=>x(x+2)=0

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

c: 2x(x-3)=x-3

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

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

=>\(\left[\begin{array}{l}x-3=0\\ 2x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac12\end{array}\right.\)

d: 2x(x-3)+5x-15=0

=>2x(x-3)+5(x-3)=0

=>(x-3)(2x+5)=0

=>\(\left[\begin{array}{l}x-3=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac52\end{array}\right.\)

e: \(\left(1+x\right)^2-\left(x-1\right)^2=0\)

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

=>2*2x=0

=>4x=0

=>x=0

f: \(\left(x-2\right)^2=\left(3x+5\right)^2\)

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

=>(3x+5+x-2)(3x+5-x+2)=0

=>(4x+3)(2x+7)=0

=>\(\left[\begin{array}{l}4x+3=0\\ 2x+7=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac34\\ x=-\frac72\end{array}\right.\)

g: \(\left(6-9x\right)^2=\left(5x-7\right)^2\)

=>\(\left(9x-6\right)^2-\left(5x-7\right)^2=0\)

=>(9x-6-5x+7)(9x-6+5x-7)=0

=>(4x+1)(14x-13)=0

=>\(\left[\begin{array}{l}4x+1=0\\ 14x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac14\\ x=\frac{13}{14}\end{array}\right.\)

h: \(\left(x+1\right)^2\cdot\left(x+2\right)=0\)

=>\(\left[\begin{array}{l}x+1=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1\\ x=-2\end{array}\right.\)

i: \(\left(3x-1\right)\cdot\left(3-x\right)^2=0\)

=>\(\left[\begin{array}{l}3x-1=0\\ \left(3-x\right)^2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=1\\ 3-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=3\end{array}\right.\)

Bài 1:

Đặt \(\hept{\begin{cases}S=x+y\\P=xy\end{cases}}\) hpt thành:

\(\hept{\begin{cases}S^2-P=3\\S+P=9\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}S^2-P=3\\S=9-P\end{cases}}\Leftrightarrow\left(9-P\right)^2-P=3\)

\(\Leftrightarrow\orbr{\begin{cases}P=6\Rightarrow S=3\\P=13\Rightarrow S=-4\end{cases}}\).Thay 2 trường hợp S và P vào ta tìm dc

\(\hept{\begin{cases}x=3\\y=0\end{cases}}\)và\(\hept{\begin{cases}x=0\\y=3\end{cases}}\)

Câu 3: ĐK: \(x\ge0\)

Ta thấy \(x-\sqrt{x-1}=0\Rightarrow x=\sqrt{x-1}\Rightarrow x^2-x+1=0\) (Vô lý), vì thế \(x-\sqrt{x-1}\ne0.\)

Khi đó \(pt\Leftrightarrow\frac{3\left[x^2-\left(x-1\right)\right]}{x+\sqrt{x-1}}=x+\sqrt{x-1}\Rightarrow3\left(x-\sqrt{x-1}\right)=x+\sqrt{x-1}\)

\(\Rightarrow2x-4\sqrt{x-1}=0\)

Đặt \(\sqrt{x-1}=t\Rightarrow x=t^2+1\Rightarrow2\left(t^2+1\right)-4t=0\Rightarrow t=1\Rightarrow x=2\left(tm\right)\)

1, \(x^2-5x+4-\sqrt{5-x}-\sqrt{x-2}=0\)ĐKXĐ \(2\le x\le5\)

ĐK dấu bằng xảy ra \(x^2-5x+4\ge0\)

Kết hơp với ĐKXĐ=> \(4\le x\le5\)

Khi đó Phương trình tương đương

\(x^2-7x+11+\left(x-4-\sqrt{5-x}\right)+\left(x-3-\sqrt{x-2}\right)=0\)

<=> \(x^2-7x+11+\frac{x^2-7x+11}{x-4+\sqrt{5-x}}+\frac{x^2-7x+11}{x-3+\sqrt{x-2}}=0\)

=> \(\orbr{\begin{cases}x^2-7x+11=0\\1+\frac{1}{x-4+\sqrt{5-x}}+\frac{1}{x-3+\sqrt{x-2}}=0\left(2\right)\end{cases}}\)

Phương trình (2) vô nghiệm với \(4\le x\le5\)=> VT>0

\(x^2-7x+11=0\)

Với \(4\le x\le5\)

\(S=\left\{\frac{7+\sqrt{5}}{2}\right\}\)

2.\(\sqrt{x+2}+\sqrt{3-x}=x^3+x^2-4x-1\)ĐKXĐ \(-2\le x\le3\)

<=> \(3x^3+3x^2-12x-3=3\sqrt{x+2}+3\sqrt{3-x}\)

<=> \(3x^3+3x^2-12x-12+\left(x+4-3\sqrt{x+2}\right)+\left(5-x-3\sqrt{3-x}\right)=0\)

<=> \(3\left(x^2-x-2\right)\left(x+2\right)+\frac{x^2-x-2}{x+4+3\sqrt{x+2}}+\frac{x^2-x-2}{5-x+3\sqrt{3-x}}=0\)

=> \(\orbr{\begin{cases}x^2-x-2=0\\3\left(x+2\right)+\frac{1}{x+4+3\sqrt{x+2}}+\frac{1}{5-x+3\sqrt{x-3}}=0\left(2\right)\end{cases}}\)

Phương trình (2) vô nghiệm với\(-2\le x\le3\)=> VT>0

\(S=\left\{2;-1\right\}\)

Bài 1 :

a) \(x^3-x^2-x-2=0\)

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

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

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

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

Vì \(x^2+x+1=x^2+2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)

Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow x^2+x+1\ge\frac{3}{4}\forall x\)(2)

Từ (1) và (2) \(\Rightarrow x-2=0\)\(\Leftrightarrow x=2\)

Vậy \(x=2\)

Bài 2: 

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

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

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

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

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

Vì \(\left(x-y-1\right)^2\ge0\forall x,y\); \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-y-1\right)^2+\left(x-2\right)^2\ge0\forall x,y\)(2)

Từ (1) và (2) \(\Rightarrow\left(x-y-1\right)^2+\left(x-y\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}x-y-1=0\\x-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=x-1\\x=2\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=2\end{cases}}\)

Vậy \(x=2\)và \(y=1\)

ĐK: x khác 0

\(\hept{\begin{cases}x\left(x+y\right)+x=3\\x^2\left(x+y\right)^2+x^2=5\end{cases}}\)

Đặt: x(x+y)=u, x=v

Ta có hệ mới :

\(\hept{\begin{cases}u+v=3\\u^2+v^2=5\end{cases}}\)Hệ phương trình đối xứng loại 1, em làm tiếp nhé!

554554544