2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)

\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)

Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)

\(\Rightarrow x=3\)

c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)

\(\Rightarrow x=1\)

\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)

bạn làm nốt pần này nhá

1. ĐK: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{3x-2}\ge0\\b=\sqrt{x-1}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}ab=\sqrt{\left(3x-2\right)\left(x-1\right)}=\sqrt{3x^2-5x+2}\\a^2+b^2=\left(3x-2\right)+\left(x-1\right)=4x-3\end{matrix}\right.\)

pt trên được viết lại thành

\(a+b=a^2+b^2-6+2ab\)

\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-6=0\)

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

\(\Leftrightarrow a+b=3\) (vì \(a,b\ge0\))

\(\Rightarrow\sqrt{3x-2}+\sqrt{x-1}=3\)

Đến đây thì dễ rồi, bạn bình phương 2 lần để tìm x, sau đó đối chiếu với ĐK để loại nghiệm.

2. ĐK: \(-\sqrt{17}\le x\le\sqrt{17}\)

Đặt \(\left\{{}\begin{matrix}a=x\\b=\sqrt{17-x^2}\ge0\end{matrix}\right.\)

Ta lập được hệ phương trình

\(\left\{{}\begin{matrix}a+b+ab=9\\a^2+b^2=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+ab=9\\\left(a+b\right)^2-2ab=17\end{matrix}\right.\) (I)

Đặt S=x+y; P=xy thì

\(\left(I\right)\Rightarrow\left\{{}\begin{matrix}S+P=9\\S^2-2P=17\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}S=5\\P=4\end{matrix}\right.\\\left\{{}\begin{matrix}S=-7\\P=16\end{matrix}\right.\end{matrix}\right.\)

Đến đây dễ rồi bạn làm tiếp nha

Em xin phép làm bài EZ nhất :)

4,ĐK :\(\forall x\in R\)

Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))

\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)

\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)

\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy ....

a)...ghi lại đề...

\(\Leftrightarrow\sqrt{x^2-x-2x+2}=\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{x\left(x-1\right)-2\left(x-1\right)}=\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{x-2}\cdot\sqrt{x-1}=\sqrt{x-1}\)

\(\Leftrightarrow\sqrt{x-2}=\frac{\sqrt{x-1}}{\sqrt{x-1}}=1\)

\(\Leftrightarrow\sqrt{x-2}^2=1^2\)

\(\Leftrightarrow x-2=1\)(Vì \(x-2\ge0\Leftrightarrow x\ge2\))

\(\Leftrightarrow x=3\)

\(\)

\(a,\sqrt{x^2-3x+2}=\sqrt{x-1}\)

\(\Rightarrow x^2-3x+2=x-1\)

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

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

\(\Rightarrow\left(x-3\right)\left(x-1\right)=0\)

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

Vậy..........

vô đây Câu hỏi của Phan hữu Dũng - Toán lớp 9 - Học toán với OnlineMath

Cho mình copy nhé:

Đặt \(\sqrt{3x-2}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(\Rightarrow\begin{cases}a^2=3x-2\\b^2=x-1\end{cases}\)\(\Rightarrow a^2+b^2=4x-3\)

\(pt\Leftrightarrow a+b=a^2+b^2-6+2ab\)

\(\Leftrightarrow a^2+b^2-6+2ab-a-b=0\)

\(\Leftrightarrow\left(a+b\right)^2-\left(a+b\right)-6=0\)

\(\Leftrightarrow\left(a+b\right)^2+2\left(a+b\right)-3\left(a+b\right)-6=0\)

\(\Leftrightarrow\left(a+b\right)\left(a+b+2\right)-3\left(a+b+2\right)=0\)

\(\Leftrightarrow\left(a+b-3\right)\left(a+b+2\right)=0\)

\(\Leftrightarrow a+b=3\)hoặc\(a+b=-2\)(loại,vì a\(\ge\)0;b\(\ge\)0 =>a+b\(\ge\)0)

• Với a+b=3

\(\Rightarrow\sqrt{3x-2}+\sqrt{x-1}=3\)

\(\Leftrightarrow\sqrt{3x-2}=3-\sqrt{x-1}\)

\(\Rightarrow3x-2=9+x-1-6\sqrt{x-1}\)

\(\Rightarrow2x-10=-6\sqrt{x-1}\)

\(\Rightarrow4x^2-40x+100=36\left(x-1\right)\)

\(\Rightarrow4x^2-76x+1236=0\)

\(\Rightarrow4x^2-8x-68x+136=0\)

\(\Rightarrow4x\left(x-2\right)-68\left(x-2\right)=0\)

\(\Rightarrow\left(4x-68\right)\left(x-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=17\left(loai\right)\\x=2\left(TM\right)\end{array}\right.\)

Vậy phương trình đã cho có nghiệm là x=2

a/ Giải rồi

b/ ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)

Pt trở thành:

\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)

\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)

\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)

\(\Leftrightarrow...\)

e/ ĐKXD: \(x>0\)

\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)

\(\Rightarrow t^2=x+\frac{1}{4x}+1\)

Pt trở thành:

\(5t=2\left(t^2-1\right)+4\)

\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)

\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)

a)

ĐK: $x\geq 2$

PT \(\Leftrightarrow \sqrt{(x-1)(x-2)}=\sqrt{x-1}\)

\(\Leftrightarrow \sqrt{x-1}(\sqrt{x-2}-1)=0\)

\(\Rightarrow \left[\begin{matrix} \sqrt{x-1}=0\\ \sqrt{x-2}-1=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1(\text{loại vì x}\geq 2)\\ \sqrt{x-2}=1\end{matrix}\right.\)

\(\Rightarrow x=1^2+2=3\) là nghiệm duy nhất thỏa mãn

b)

ĐK: $x\in\mathbb{R}$

Bình phương 2 vế:

\(\Rightarrow x^2-4x+4=4x^2-12x+9\)

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

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

\(\Leftrightarrow (-x+1)(3x-5)=0\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{5}{3}\end{matrix}\right.\) (đều thỏa mãn)

Vậy..........

c)

ĐKXĐ: $x\geq 3$

PT \(\Leftrightarrow \sqrt{(x-2)(x-3)}=\sqrt{x-2}\)

\(\Leftrightarrow (x-2)(x-3)=x-2\) (bình phương 2 vế không âm)

\(\Leftrightarrow (x-2)(x-3-1)=0\)

\(\Rightarrow \left[\begin{matrix} x-2=0\\ x-4=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2(\text{loại vì x}\geq 3)\\ x=4\end{matrix}\right.\)

Vậy $x=4$

d)

ĐK: $x\in\mathbb{R}$

PT \(\Leftrightarrow 4x^2-4x+1=x^2-6x+9\) (bình phương 2 vế không âm)

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

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

\(\Leftrightarrow (x+2)(3x-4)=0\Rightarrow \left[\begin{matrix} x+2=0\\ 3x-4=0\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=-2\\ x=\frac{4}{3}\end{matrix}\right.\) (đều thỏa mãn)

Vậy.........