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Bài 1: ĐKXĐ: $2\leq x\leq 4$
PT $\Leftrightarrow (\sqrt{x-2}+\sqrt{4-x})^2=2$
$\Leftrightarrow 2+2\sqrt{(x-2)(4-x)}=2$
$\Leftrightarrow (x-2)(4-x)=0$
$\Leftrightarrow x-2=0$ hoặc $4-x=0$
$\Leftrightarrow x=2$ hoặc $x=4$ (tm)
Bài 2:
PT $\Leftrightarrow 4x^3(x-1)-3x^2(x-1)+6x(x-1)-4(x-1)=0$
$\Leftrightarrow (x-1)(4x^3-3x^2+6x-4)=0$
$\Leftrightarrow x=1$ hoặc $4x^3-3x^2+6x-4=0$
Với $4x^3-3x^2+6x-4=0(*)$
Đặt $x=t+\frac{1}{4}$ thì pt $(*)$ trở thành:
$4t^3+\frac{21}{4}t-\frac{21}{8}=0$
Đặt $t=m-\frac{7}{16m}$ thì pt trở thành:
$4m^3-\frac{343}{1024m^3}-\frac{21}{8}=0$
$\Leftrightarrow 4096m^6-2688m^3-343=0$
Coi đây là pt bậc 2 ẩn $m^3$ và giải ta thu được \(m=\frac{\sqrt[3]{49}}{4}\) hoặc \(m=\frac{-\sqrt[3]{7}}{4}\)
Khi đó ta thu được \(x=\frac{1}{4}(1-\sqrt[3]{7}+\sqrt[3]{49})\)
Giải các phương trình sau
a)\(x^2-2-x+\sqrt{2}=0\)
b) \((1-\sqrt{2})x^2-2(1+\sqrt{2})x+1+3\sqrt{2}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^2-2-x+\sqrt{2}=0\)
=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)=0\)
=>\(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-1\right)=0\)
=>\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}+1\end{matrix}\right.\)
b: \(\left(1-\sqrt{2}\right)x^2-2\left(1+\sqrt{2}\right)x+1+3\sqrt{2}=0\)
\(\Delta=\left(-2-2\sqrt{2}\right)^2-4\left(1-\sqrt{2}\right)\left(1+3\sqrt{2}\right)\)
\(=12+8\sqrt{2}+4\left(\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)\)
\(=12+8\sqrt{2}+4\left(6+\sqrt{2}-3\sqrt{2}-1\right)\)
\(=12+8\sqrt{2}+24-8\sqrt{2}-4=32>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{2\left(1+\sqrt{2}\right)-4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=1\\x_2=\dfrac{2\left(1+\sqrt{2}\right)+4\sqrt{2}}{2\left(1-\sqrt{2}\right)}=-7-4\sqrt{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2+3x-5=0\)
\(< =>2x^2-2x+5x-5=0\)
\(< =>2x\left(x-1\right)+5\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(2x+5\right)=0\)
\(< =>\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
\(\hept{\begin{cases}x+2y=1\\-3x+4y=-18\end{cases}}\)
\(< =>\hept{\begin{cases}-3x-6y=-3\\-3x-6y+10y=-18\end{cases}}\)
\(< =>\hept{\begin{cases}x+2y=1\\10y=-18+3=-15\end{cases}}\)
\(< =>\hept{\begin{cases}x+2y=1\\y=-\frac{3}{2}\end{cases}< =>\hept{\begin{cases}x-3=1\\y=-\frac{3}{2}\end{cases}< =>\hept{\begin{cases}x=4\\y=-\frac{3}{2}\end{cases}}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ĐKXĐ: \(x^2-1\ge0\)
Đặt \(\sqrt{x^2-1}=t\left(t\ge0\right)\)
\(\Rightarrow t=t^2\Rightarrow t\left(t-1\right)=0\Rightarrow\left[{}\begin{matrix}t=0\\t=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-1}=0\\\sqrt{x^2-1}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm1\\x=\pm\sqrt{2}\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge2\)
Ta có: \(\sqrt{x-2}+\sqrt{x-3}\ge0\) mà \(\sqrt{x-2}+\sqrt{x-3}=-5< 0\Rightarrow\) không có x thỏa
c) \(\sqrt{x^2+4x+4}+\left|x-4\right|=0\)
\(\Rightarrow\left|x+2\right|+\left|x-4\right|=0\) mà \(\left|x+2\right|+\left|x-4\right|\ge0\Rightarrow\left\{{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\)
\(\Rightarrow\) không có x thỏa
![](https://rs.olm.vn/images/avt/0.png?1311)
a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)
`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`
`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\)
`b)sqrt{x-3}/sqrt{2x+1}=2`
ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)
`<=>x>=3`
`pt<=>sqrt{x-3}=2sqrt{2x+1}`
`<=>x-3=8x+4`
`<=>7x=7`
`<=>x=1(l)`
`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`
`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`
`<=>|x-1|+|x-2|=3`
`**x>=2`
`pt<=>x-1+x-2=3`
`<=>2x=6`
`<=>x=3(tm)`
`**x<=1`
`pt<=>1-x+2-x=3`
`<=>3-x=3`
`<=>x=0(tm)`
`**1<=x<=2`
`pt<=>x-1+2-x=3`
`<=>=-1=3` vô lý
Vậy `S={0,3}`
![](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)
1) \(\sqrt[]{9\left(x-1\right)}=21\)
\(\Leftrightarrow9\left(x-1\right)=21^2\)
\(\Leftrightarrow9\left(x-1\right)=441\)
\(\Leftrightarrow x-1=49\Leftrightarrow x=50\)
2) \(\sqrt[]{1-x}+\sqrt[]{4-4x}-\dfrac{1}{3}\sqrt[]{16-16x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}+\sqrt[]{4\left(1-x\right)}-\dfrac{1}{3}\sqrt[]{16\left(1-x\right)}+5=0\)
\(\)\(\Leftrightarrow\sqrt[]{1-x}+2\sqrt[]{1-x}-\dfrac{4}{3}\sqrt[]{1-x}+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}\left(1+3-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\sqrt[]{1-x}.\dfrac{8}{3}=-5\)
\(\Leftrightarrow\sqrt[]{1-x}=-\dfrac{15}{8}\)
mà \(\sqrt[]{1-x}\ge0\)
\(\Leftrightarrow pt.vô.nghiệm\)
3) \(\sqrt[]{2x}-\sqrt[]{50}=0\)
\(\Leftrightarrow\sqrt[]{2x}=\sqrt[]{50}\)
\(\Leftrightarrow2x=50\Leftrightarrow x=25\)
1) \(\sqrt{9\left(x-1\right)}=21\) (ĐK: \(x\ge1\))
\(\Leftrightarrow3\sqrt{x-1}=21\)
\(\Leftrightarrow\sqrt{x-1}=7\)
\(\Leftrightarrow x-1=49\)
\(\Leftrightarrow x=49+1\)
\(\Leftrightarrow x=50\left(tm\right)\)
2) \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\) (ĐK: \(x\le1\))
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{1-x}-\dfrac{4}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}=-5\) (vô lý)
Phương trình vô nghiệm
3) \(\sqrt{2x}-\sqrt{50}=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow\sqrt{2x}=\sqrt{50}\)
\(\Leftrightarrow2x=50\)
\(\Leftrightarrow x=\dfrac{50}{2}\)
\(\Leftrightarrow x=25\left(tm\right)\)
4) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\left(ĐK:x\ge-\dfrac{1}{2}\right)\\2x+1=-6\left(ĐK:x< -\dfrac{1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(tm\right)\\x=-\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
5) \(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow\left|x-3\right|=3-x\)
\(\Leftrightarrow x-3=3-x\)
\(\Leftrightarrow x+x=3+3\)
\(\Leftrightarrow x=\dfrac{6}{2}\)
\(\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Delta'=\left(\sqrt{3}+1\right)^2-2\sqrt{3}=4>0\)
\(\Rightarrow\) Phương trình đã cho có 2 nghiệm pb:
\(\left[{}\begin{matrix}x=-\sqrt{3}-1-\sqrt{4}=-\sqrt{3}-3\\x=-\sqrt{3}-1+\sqrt{4}=-\sqrt{3}+1\end{matrix}\right.\)