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![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-2-2\sqrt{4x-7}=0\)
\(\Leftrightarrow\left(4x-7-2\sqrt{4x-7}+1\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(\sqrt{4x-7}-1\right)^2+\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{4x-7}-1=0\\x-2=0\end{matrix}\right.\)
Tự làm tiếp nhé.
. . .
\(4x^2-5x+1+2\sqrt{x-1}=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-1\right)+2\sqrt{x-1}=0\)
\(\Leftrightarrow\sqrt{x-1}\left[\left(4x-1\right)\sqrt{x-1}+2\right]=0\)
\(\Rightarrow x=1\)
. . .
\(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
\(\Leftrightarrow\sqrt{\left(x-2\right)^2}+\sqrt{\left(x-3\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|+\left|x-3\right|=1\)
\(VT=\left|x-2\right|+\left|3-x\right|\ge\left|x-2+3-x\right|=1=VP\)
Dấu "=" xảy ra khi \(\left(x-2\right)\left(3-x\right)\ge0\)
Đến đây lập bảng xét dấu
. . .
\(x^2-x+2=2\sqrt{x^2-x+1}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+1}-1\right)^2=0\)
Tự làm tiếp nhé.
\(\sqrt{3x+1}-\sqrt{6-x}+3x^2-14x-8=0\)
\(\Leftrightarrow\left(\sqrt{3x+1}-4\right)+\left(1-\sqrt{6-x}\right)+\left(3x^2-14-5\right)=0\)
\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{1-6+x}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{x-5}{1+\sqrt{6-x}}+\left(x-5\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left(\dfrac{3}{\sqrt{3x+1}+4}+\dfrac{1}{1+\sqrt{6-x}}+3x+1\right)\left(x-5\right)=0\)
\(\Rightarrow x=5\)
. . .
\(\sqrt{2x^2-4x+5}-x+4=0\)
\(\Leftrightarrow\sqrt{2x^2-4x+5}=x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\2x^2-4x+5=x^2-8x+16\end{matrix}\right.\)
Tự làm tiếp nhé.
. . .
\(\sqrt{2x+3}+\sqrt{x-1}=\sqrt{x+6}\)
\(\Leftrightarrow\sqrt{2x+3}=\sqrt{x+6}-\sqrt{x-1}\)
\(\Leftrightarrow2x+3=x+6-2\sqrt{\left(x+6\right)\left(x-1\right)}+x-1\)
\(\Leftrightarrow2\sqrt{x^2+5x-6}=2\)
\(\Leftrightarrow x^2+5x-6=1\)
Tự làm tiếp nhé.
. . .
\(x+y+\dfrac{1}{2}=\sqrt{x}+\sqrt{y}\)
\(\Leftrightarrow\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\left(y-\sqrt{y}+\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\left(\sqrt{y}-\dfrac{1}{2}\right)^2=0\)
Tự làm tiếp nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)
\(\Rightarrow3x-2\sqrt{x-1}-4=0\)
\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)
\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)
\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)
*TH1: x = 2 (t/m)
*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)
\(\Rightarrow3\sqrt{x-1}+3=2\)
\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)
Vậy S = {2}
b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )
\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)
\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)
\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)
=> x = 2
\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)
\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: x>=3
Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)
=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)
=>\(\dfrac{3}{2}\sqrt{x-3}=3\)
=>\(\sqrt{x-3}=2\)
=>x-3=4
=>x=7(nhận)
b: ĐKXĐ: x>=0
\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)
=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)
=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)
=>\(7\sqrt{x}-5< =0\)
=>\(\sqrt{x}< =\dfrac{5}{7}\)
=>0<=x<=25/49
c: ĐKXĐ: x>=5
\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)
=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)
=>\(\dfrac{3}{2}\sqrt{x-5}=3\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đk: \(x\ge1\)
\(\Leftrightarrow4\left(2\sqrt{x-1}-1\right)+\left(4x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\dfrac{4\left(4x-5\right)}{2\sqrt{x-1}+1}+\left(4x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(4x-5\right)\left(\dfrac{4}{2\sqrt{x-1}+1}+x+2\right)=0\)
\(\Leftrightarrow x=\dfrac{5}{4}\)(Dễ thấy ngoặc to lớn hơn 0 với \(x\ge1\))
![](https://rs.olm.vn/images/avt/0.png?1311)
1.
\(\sqrt{50}-3\sqrt{8}+\sqrt{32}=5\sqrt{2}-6\sqrt{2}+4\sqrt{2}=3\sqrt{2}\)
2.
a, ĐK: \(x\in R\)
\(pt\Leftrightarrow\sqrt{\left(x-2\right)^2}=1\)
\(\Leftrightarrow\left|x-2\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
b, ĐK: \(x\ge3\)
\(pt\Leftrightarrow\sqrt{x-3}\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=1\left(l\right)\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)√x2−9 - 3√x−3 =0
<=> (√x-3)(√x+3)-3√x-3=0
<=> (√x-3)(√x+3-3)=0
<=> (√x-3)√x=0
<=> √x-3=0
<=>x=9
b)√4x2−12x+9=x - 3
<=> √(2x -3)2 =x-3
<=> 2x-3=x-3
<=>2x-x=-3+3
<=>x=0
c)√x2+6x+9=3x-1
<=> √(x+3)2 =3x-1
<=> x+3=3x-1
<=> -2x=-4
<=> x=2
Nhớ cho mình 1 tim nha bạn
Sau em nên gõ các kí hiệu toán học ở phần Σ để mọi người dễ dàng đọc hơn nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
f) Ta có: \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
\(\Leftrightarrow4\left|x+1\right|-3\left|x+1\right|=4\)
\(\Leftrightarrow\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g) Ta có: \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow5\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
\(a,Đk:x\ge0\\ PT\Leftrightarrow4x-8\sqrt{x}-7\sqrt{x}+14=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(4\sqrt{x}-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{49}{4}\end{matrix}\right.\left(tm\right)\\ b,ĐK:x\ge0\\ PT\Leftrightarrow\sqrt{x+1}-\sqrt{3x}+1-4x^2=0\\ \Leftrightarrow\dfrac{1-2x}{\sqrt{x+1}+\sqrt{3x}}+\left(1-2x\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\\dfrac{1}{\sqrt{x+1}+\sqrt{3x}}+2x+1=0\left(1\right)\end{matrix}\right.\)
Với \(x\ge0\Leftrightarrow\left(1\right)>0\)
Vậy PT có nghiệm \(x=\dfrac{1}{2}\)