Đề bài: Giải phương trình sau trên tập số thực:

\(\sqrt{5x^{2}-14x+9}-\sqrt{x^{2}-x-20}=5\sqrt{x+1}\)

Bài giải: Điều kiện \(x\geqslant 5\)

Chuyển vế và bình phương hai vế phương trình ta có

\(2x^{2}-5x+2=5\sqrt{\left ( x^{2}-x-20 \right )\left ( x+1 \right )}\)

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

Ta cần tìm các hằng số \(a,b\) sao cho

\(a\left ( x^{2}-4x-5 \right )+b\left ( x+4 \right )=2x^{2}-5x+2\)

Đồng nhất hai vế đẳng thức trên ta có hệ phương trình

\(\left\{\begin{matrix} a=2 & & \\ -4a+b=-5 & & \\ -5a+4b=2 & & \end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} a=2 & & \\ b=3 & & \end{matrix}\right.\)

Đặt \(u=\sqrt{x^{2}-4x-5}; v=\sqrt{x+4}\), ta có phương trình

\(2a^{2}+3b^{2}=5ab\Leftrightarrow \left ( a-b \right )\left ( 2a-3b \right )=0\)

TH1: \(a=b\) thì \(x=\frac{5+\sqrt{61}}{2}\)

TH2: \(2a=3b\) thì \(x=8\)

Vậy nghiệm của phương trình là \(x=8;x=\frac{5+\sqrt{61}}{2}\)

a) \(\left(x+2\right).\left(x+3\right)-\left(x-2\right).\left(x+5\right)=6\)

\(\Leftrightarrow\left(x+2\right).x+\left(x+2\right).3-x.\left(x+5\right)+2.\left(x+5\right)=6\)

\(\Leftrightarrow x^2+2x+3x+6-x^2-5x+2x+10=6\)

\(\Leftrightarrow2x+16=6\)

\(\Leftrightarrow2x=-10\)

\(\Leftrightarrow x=-5\)

Tìm x,biết:

a/\(x+5x^2=0\Leftrightarrow......\)
b/\(x+1=\left(x+1\right)^2\Leftrightarrow..........\)
c/\(x^3+x=0\Leftrightarrow.......\)
d/\(5x\left(x-2\right)-\left(2-x\right)=0\)
e/\(x\left(2x-1\right)+\frac{1}{3}-\frac{2}{3}x=0\Leftrightarrow........\)
g/\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow.....\)
h/\(x^2-3x=0\Leftrightarrow.....\)
i/\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow.....\)

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)

\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)

\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0

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

\(\Leftrightarrow x^2+2x-10x-20=0\)

\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0

\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0

\(\Leftrightarrow\)

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)

\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)

\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)

\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0

\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0

\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0

\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

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

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0

\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0

\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0

\(\Leftrightarrow\) 2x = 8 hoặc x = 1

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)

Vậy S = {4; 1}

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

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

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

\(\Leftrightarrow\) 4 (x - 1) =0

\(\Leftrightarrow\) x - 1 = 0 / 4 = 0

\(\Leftrightarrow\) x = 1 (Nhận)

Vậy S = {1}

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)

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

\(\Leftrightarrow\) 0

Vậy S ={\(\varnothing\)}

Giải các bất phương trình, hệ phương trình

a) \(\dfrac{x^2\left(3x-2\right)\left(x^2-1\right)}{\left(-x^2+2x-3\right)\left(2-x\right)^2}\ge0\)

b) \(\dfrac{x-5}{x-1}>2\)

c) \(2x-\sqrt{x^2-5x-14}< 1\)

d) \(x+\sqrt{x^2-4x-5}< 4\)

e) \(\left\{{}\begin{matrix}\left(4-x\right)\left(x^2-2x-3\right)< 0\\x^2\ge\left(x^2-x-3\right)^2\end{matrix}\right.\)

\(\left(x^2-x+1\right)^4-6x^2\left(x^2-x+1\right)^2+5x^4=0\)

\(\Leftrightarrow\left[\left(x^2-x+1\right)^2\right]^2-2\left(x^2-x+1\right)^2.3x^2+\left(3x^2\right)^2-4x^4=0\)

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

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

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

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

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

Mấy bạn cho mình gửi tạm nha, xíu mình nhờ CTV xóa :(