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![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(x^2-4\sqrt{15}x+19=0\\ < =>x^2-4\sqrt{15}x+60-41=0\\ < =>\left(x-2\sqrt{15}\right)^2-41=0\\ < =>\left(x-2\sqrt{15}-\sqrt{41}\right)\left(x-2\sqrt{15}+\sqrt{41}\right)=0\\ < =>\left[{}\begin{matrix}x-2\sqrt{15}-\sqrt{41}=0\\x-2\sqrt{15}+\sqrt{41}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=2\sqrt{15}+\sqrt{41}\\x=2\sqrt{15}-\sqrt{41}\end{matrix}\right.\)
b)
\(4x^2+4\sqrt{5}x+5=0\\ < =>\left(2x+\sqrt{5}\right)^2=0\\ < =>2x+\sqrt{5}=0\\ < =>2x=-\sqrt{5}\\ < =>-\dfrac{\sqrt{5}}{2}\)
a: Δ=(4căn 15)^2-4*1*19=164>0
Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x=\dfrac{4\sqrt{5}-2\sqrt{41}}{2}=2\sqrt{5}-\sqrt{41}\\x_2=2\sqrt{5}+\sqrt{41}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot\sqrt{5}+5=0\)
=>(2x+căn 5)^2=0
=>2x+căn 5=0
=>x=-1/2*căn 5
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: x ≥ -3
Phương trình tương đương:
4√(x + 3) + √(x + 3) = 15
⇔ 5√(x + 3) = 15
⇔ √(x + 3) = 15 : 3
⇔ √(x + 3) = 3
⇔ x + 3 = 9
⇔ x = 9 - 3
⇔ x = 6 (nhận)
Vậy S = {6}
b) ĐKXĐ: x ≥ 2
Phương trình tương đương:
√[(x - 2)(x + 2)] - 3√(x - 2) = 0
⇔ √(x - 2)√(x + 2 - 3) = 0
⇔ √(x - 2)√(x - 1) = 0
⇔ √(x - 2) = 0 hoặc √(x - 1) = 0
*) √(x - 2) = 0
⇔ x - 2 = 0
⇔ x = 2 (nhận)
*) √(x - 1) = 0
⇔ x - 1 = 0
⇔ x = 1 (loại)
Vậy S = {2}
![](https://rs.olm.vn/images/avt/0.png?1311)
a: ĐKXĐ: \(\left\{{}\begin{matrix}2x-3>=0\\x-1>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{3}{2}\\x>1\end{matrix}\right.\Leftrightarrow x>=\dfrac{3}{2}\)
\(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
=>\(\sqrt{\dfrac{2x-3}{x-1}}=2\)
=>\(\dfrac{2x-3}{x-1}=4\)
=>4(x-1)=2x-3
=>4x-4=2x-3
=>4x-2x=-3+4
=>2x=1
=>\(x=\dfrac{1}{2}\left(loại\right)\)
b: ĐKXĐ: 2x+15>=0
=>x>=-15/2
\(x+\sqrt{2x+15}=0\)
=>\(\sqrt{2x+5}=-x\)
=>\(\left\{{}\begin{matrix}-x>=0\\\left(-x\right)^2=2x+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\x^2-2x-5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left(x-1\right)^2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left[{}\begin{matrix}x-1=\sqrt{6}\\x-1=-\sqrt{6}\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-\dfrac{15}{2}< =x< =0\\\left[{}\begin{matrix}x=\sqrt{6}+1\left(loại\right)\\x=-\sqrt{6}+1\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
2:
\(A=\dfrac{x_2-1+x_1-1}{x_1x_2-\left(x_1+x_2\right)+1}\)
\(=\dfrac{3-2}{-7-3+1}=\dfrac{1}{-9}=\dfrac{-1}{9}\)
B=(x1+x2)^2-2x1x2
=3^2-2*(-7)
=9+14=23
C=căn (x1+x2)^2-4x1x2
=căn 3^2-4*(-7)=căn 9+28=căn 27
D=(x1^2+x2^2)^2-2(x1x2)^2
=23^2-2*(-7)^2
=23^2-2*49=431
D=9x1x2+3(x1^2+x2^2)+x1x2
=10x1x2+3*23
=69+10*(-7)=-1
![](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)
????
xin lỗi nha !
mình mới học lớp 3
mà bài này khó nắm
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ ĐK: \(x \ge -1\). Đặt \(\sqrt{x+1}=a \ge 0\)
PT: \(\Leftrightarrow6a-3a-2a=5\)
\(\Leftrightarrow a=5\)
\(\Leftrightarrow x+1=15\Leftrightarrow x=24\) (nhận)
b,c: Hai ý này đều làm theo cách bình phương hoặc đưa về phương trình chứa dấu giá trị tuyệt đối được nhé.
b) Cách 1: ĐKXĐ: Tự tìm
\(\sqrt{x^{2}-4x+4}=2\Leftrightarrow x^{2}-4x+4=4\Leftrightarrow x(x-4)=0\)
\(\Leftrightarrow x=0\) hoặc \(x=4\) cả 2 cái này đều TMĐK
Cách 2: \((\sqrt{x^2-4x+4}=2)\)
\(\Leftrightarrow \sqrt{(x-2)^2}=2\)
\(\Leftrightarrow \mid x-2\mid=2\)
Với \(x\geq 2\) thì :
\(x-2=2 \Leftrightarrow x=4\) (nhận)
Với \(x<2\) thì
\(-x-2=2\Leftrightarrow x=0\) (nhận)
Vậy \(S={0;4}\)
c) Cách 1: \(\sqrt{x^{2}-6x+9}=x-2\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x^{2}-6x+9=x^{2}-4x+4 \end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix}x\geq 2 \\ x=\frac{5}{2} \end{matrix}\right.\)
Nghiệm TMĐK
Cách 2: \((\sqrt{x^2-6x+9}=x-2)\)
\(\Leftrightarrow \mid x-3\mid =x-2\)
Với \(x\geq 3\) thì
\(x-3=x-2\Leftrightarrow 0x=-1\) ( vô lý)
Với \(x<3\) thì
\(-x+3=x-2\Leftrightarrow -2x=-5 \Leftrightarrow x=\frac{5}{2}\)
Vậy \(S={\frac{5}{2}}\)
d) ĐKXĐ: Tự tìm
\(\sqrt{x^{2}+4}=\sqrt{2x+3}\Leftrightarrow x^{2}+4=2x+3\Leftrightarrow x^{2}-2x+1=0\Leftrightarrow (x-1)^{2}=0\)
\(\Leftrightarrow x=1\)
e) ĐKXĐ: \(x\geq \frac{3}{2}\)
\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\Leftrightarrow \frac{2x-3}{x-1}=4\Rightarrow 2x-3=4x-4\Leftrightarrow x=\frac{1}{2}\)
Nghiệm không TMĐK.
Phương trình vô nghiệm.
f) ĐKXĐ: \(x\geq \frac{-15}{2}\)
\(x+\sqrt{2x+15}=0\Leftrightarrow 2x+2\sqrt{2x+15}=0\Leftrightarrow 2x+15+2\sqrt{2x+15}+1-16=0\)
\(\Leftrightarrow (\sqrt{2x+15}+1)^{2}-4^{2}=0\Leftrightarrow (\sqrt{2x+15}+5)(\sqrt{2x+15}-3)=0\)
\(\Leftrightarrow \sqrt{2x+15}-3=0\Leftrightarrow \sqrt{2x+15}=3\Leftrightarrow 2x+15=9\Leftrightarrow x=-3\) (TMĐK)
a: =>\(x\cdot\left(\sqrt{3}-1\right)=16\)
=>\(x=\dfrac{16}{\sqrt{3}-1}=8\left(\sqrt{3}+1\right)\)
b: =>(x-căn 15)^2=0
=>x-căn 15=0
=>x=căn 15