Tìm x biết:
a)\(\sqrt{4x-8}+5\sqrt{x-2}\)-\(\sqrt{9x-18}=20\)
b)\(5\sqrt{x-1}\)-\(\sqrt{36x-6}\)+\(\sqrt{9x-9}\)=\(\sqrt{8x+12}\)
Tìm x biết:
a.\(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\)
b.\(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\)
c.\(\sqrt{\left(x-2\right)^2}=10\)
d.\(\sqrt{9x^2-6x+1}=15\)
e.\(\sqrt{3x+4}=3x-8\)
c) \(\sqrt{\left(x-2\right)^2}=10\)
\(x-2=10\)
\(x=12\)
d) \(\sqrt{9x^2-6x+1}=15\)
\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)
\(\sqrt{\left(3x-1\right)^2}=15\)
\(3x-1=15\)
\(3x=16\)
\(x=\dfrac{16}{3}\)
a) \(đk:x\ge0\)
\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)
b) \(đk:x\ge-2\)
\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)
\(\Leftrightarrow13\sqrt{x+2}=26\)
\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)
c) \(pt\Leftrightarrow\left|x-2\right|=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)
\(\Leftrightarrow\left|3x-1\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)
e) \(đk:x\ge\dfrac{8}{3}\)
\(pt\Leftrightarrow3x+4=9x^2-48x+64\)
\(\Leftrightarrow9x^2-51x+60=0\)
\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)
a. \(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\) ĐK: \(x\ge0\)
<=> \(\sqrt{9.2x}+2\sqrt{4.2x}-3\sqrt{2x}=12\)
<=> \(3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)
<=> \(\sqrt{2x}\left(3+4-3\right)=12\)
<=> \(4\sqrt{2x}=12\)
<=> \(\sqrt{2x}=12:4\)
<=> \(\sqrt{2x}=3\)
<=> 2x = 32
<=> 2x = 9
<=> \(x=\dfrac{9}{2}\) (TM)
b. \(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\) ĐK: \(x\ge-2\)
<=> \(\sqrt{9\left(x+2\right)}+2\sqrt{36\left(x+2\right)}-\sqrt{4\left(x+2\right)}=26\)
<=> \(3\sqrt{x+2}+72\sqrt{x+2}-2\sqrt{x+2}=26\)
<=> \(\sqrt{x+2}\left(3+72-2\right)=26\)
<=> \(73\sqrt{x+2}=26\)
<=> \(\sqrt{x+2}=\dfrac{26}{73}\)
<=> x + 2 = \(\left(\dfrac{26}{73}\right)^2\)
<=> x + 2 = \(\dfrac{676}{5329}\)
<=> \(x=\dfrac{676}{5329}-2\)
<=> \(x=-1,873146932\) (TM)
c. \(\sqrt{\left(x-2\right)^2}=10\)
<=> \(\left|x-2\right|=10\)
<=> \(\left[{}\begin{matrix}x-2=10\left(x\ge2\right)\\x-2=-10\left(x< 2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=12\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)
d. \(\sqrt{9x^2-6x+1}=15\)
<=> \(\sqrt{\left(3x-1\right)^2}=15\)
<=> \(\left|3x-1\right|=15\)
<=> \(\left[{}\begin{matrix}3x-1=15\left(x\ge\dfrac{16}{3}\right)\\3x-1=-15\left(x< \dfrac{16}{3}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{16}{3}\left(TM\right)\\x=\dfrac{-14}{3}\left(TM\right)\end{matrix}\right.\)
e. \(\sqrt{3x+4}=3x-8\) ĐK: \(x\ge\dfrac{-4}{3}\)
<=> 3x + 4 = (3x - 8)2
<=> 3x + 4 = 9x2 - 48x + 64
<=> 9x2 - 3x - 48x + 64 - 4 = 0
<=> 9x2 - 51x + 60 = 0
<=> 9x2 - 36x - 15x + 60 = 0
<=> 9x(x - 4) - 15(x - 4) = 0
<=> (9x - 15)(x - 4) = 0
<=> \(\left[{}\begin{matrix}9x-15=0\\x-4=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{15}{9}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)
Bài 1: Tìm x, biết
a)\(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
b) \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
c)\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
d) \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)
\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)
\(\Leftrightarrow4\sqrt{x-3}=20\)
\(\Leftrightarrow x-3=25\)
hay x=28
b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7
a) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\)
b) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
c) \(\sqrt{x^2+6x-9}-2\sqrt{x^2-2x+1}+\sqrt{x^2}=0\)
a: =>2*căn x+5+căn x+5-1/3*3*căn x+5=4
=>2*căn(x+5)=4
=>căn (x+5)=2
=>x+5=4
=>x=-1
b: =>\(6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
=>2*căn x-1=16
=>x-1=64
=>x=65
c, \(\sqrt{\left(x-3\right)^2}-2\sqrt{\left(x-1\right)^2}+\sqrt{x^2}=0\\ \Leftrightarrow\left|x-3\right|-2\left|x-1\right|+\left|x\right|=0\left(1\right)\)
TH1: \(x\ge3\)
\(\left(1\right)\Rightarrow x-3-2x+2+x=0\\ \Leftrightarrow-1=0\left(loại\right)\)
TH2: \(2\le x< 3\)
\(\left(1\right)\Rightarrow3-x-2x+2+x=0\\ \Leftrightarrow-2x=-5\\ \Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\)
TH3: \(0\le x< 2\)
\(\left(1\right)\Rightarrow3-x+2x-2+x=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
TH4: \(x< 0\)
\(\left(1\right)\Rightarrow3-x+2x-2-x-=0\\ \Leftrightarrow1=0\left(loại\right)\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{5}{2}\right\}\)
2.tìm x
a)\(\sqrt{x^2-6x+9}\)
b)\(\sqrt{x^2-2x+1}\)
c)\(\sqrt{4x+12}-3\sqrt{x+3}+7\sqrt{9x+27}=20\)
d)\(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
a) \(\sqrt{x^2-6x+9}\)
\(=\sqrt{\left(x^2-2.x.3+3^2\right)}\)
\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x
⇒x∈\(R\)
b) \(\sqrt{x^2-2x+1}\)
\(=\sqrt{\left(x^2-2.x.1+1^2\right)}\)
\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x
⇒x∈\(R\)
Tìm x, biết:
a) \(\sqrt{x^2-2x+1}=2\)
b)\(\sqrt{x^2-1}=x\)
c) \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
d) \(x-5\sqrt{x-2}=-2\)
e) \(2x-3\sqrt{2x-1}-5=0\)
`a)sqrt{x^2-2x+1}=2`
`<=>sqrt{(x-1)^2}=2`
`<=>|x-1|=2`
`**x-1=2<=>x=3`
`**x-1=-1<=>x=-1`.
Vậy `S={3,-1}`
`b)sqrt{x^2-1}=x`
Điều kiện:\(\begin{cases}x^2-1 \ge 0\\x \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}x^2 \ge 1\\x \ge 0\\\end{cases}\)
`<=>x>=1`
`pt<=>x^2-1=x^2`
`<=>-1=0` vô lý
Vậy pt vô nghiệm
`c)sqrt{4x-20}+3sqrt{(x-5)/9}-1/3sqrt{9x-45}=4(x>=5)`
`pt<=>sqrt{4(x-5)}+sqrt{9*(x-5)/9}-sqrt{(9x-45)*1/9}=4`
`<=>2sqrt{x-5}+sqrt{x-5}-sqrt{x-5}=4`
`<=>2sqrt{x-5}=4`
`<=>sqrt{x-5}=2`
`<=>x-5=4`
`<=>x=9(tmđk)`
Vậy `S={9}.`
`d)x-5sqrt{x-2}=-2(x>=2)`
`<=>x-2-5sqrt{x-2}+4=0`
Đặt `a=sqrt{x-2}`
`pt<=>a^2-5a+4=0`
`<=>a_1=1,a_2=4`
`<=>sqrt{x-2}=1,sqrt{x-2}=4`
`<=>x_1=3,x_2=18`,
`e)2x-3sqrt{2x-1}-5=0`
`<=>2x-1-3sqrt{2x-1}-4=0`
Đặt `a=sqrt{2x-1}(a>=0)`
`pt<=>a^2-3a-4=0`
`a-b+c=0`
`<=>a_1=-1(l),a_2=4(tm)`
`<=>sqrt{2x-1}=4`
`<=>2x-1=16`
`<=>x=17/2(tm)`
Vậy `S={17/2}`
d.
ĐKXĐ: $x\geq 2$. Đặt $\sqrt{x-2}=a(a\geq 0)$ thì pt trở thành:
$a^2+2-5a=-2$
$\Leftrightarrow a^2-5a+4=0$
$\Leftrightarrow (a-1)(a-4)=0$
$\Rightarrow a=1$ hoặc $a=4$
$\Leftrightarrow \sqrt{x-2}=1$ hoặc $\sqrt{x-2}=4$
$\Leftrightarrow x=3$ hoặc $x=18$ (đều thỏa mãn)
e. ĐKXĐ: $x\geq \frac{1}{2}$
Đặt $\sqrt{2x-1}=a(a\geq 0)$ thì pt trở thành:
$a^2+1-3a-5=0$
$\Leftrightarrow a^2-3a-4=0$
$\Leftrightarrow (a+1)(a-4)=0$
Vì $a\geq 0$ nên $a=4$
$\Leftrightarrow \sqrt{2x-1}=4$
$\Leftrightarrow x=\frac{17}{2}$
a.
$\sqrt{x^2-2x+1}=2$
$\Leftrightarrow \sqrt{(x-1)^2}=2$
$\Leftrightarrow |x-1|=2$
$\Rightarrow x-1=\pm 2$
$\Leftrightarrow x=3$ hoặc $x=-1$ (đều thỏa mãn)
b. ĐKXĐ: $x\geq 1$ hoặc $x\leq -1$
PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2-1=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 1=0\end{matrix}\right.\) (vô lý)
Vậy pt vô nghiệm
c. ĐKXĐ: $x\geq 5$
PT $\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=4$
$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$
$\Leftrightarrow 2\sqrt{x-5}=4$
$\Leftrightarrow \sqrt{x-5}=2$
$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)
a) \(\sqrt{4x^2-9}=2\sqrt{x+3}\)
b) \(\sqrt{4x+20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
c) \(\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27\sqrt{\dfrac{x-1}{81}}=4\)
d)\(5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
Tìm x, biết:
b)\(5\sqrt{x-1}-\sqrt{36x-36}\)\(+\sqrt{9x-9}=\sqrt{8x+12}\)
Ta có: \(5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}\) \(\left(ĐK:x\ge1\right)\)
\(\Leftrightarrow5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=\sqrt{8x+12}\)
\(\Leftrightarrow2\sqrt{x-1}=\sqrt{8x+12}\)
\(\Leftrightarrow\left(2\sqrt{x-1}\right)^2=\left(\sqrt{8x+12}\right)^2\)
\(\Leftrightarrow4.\left(x-1\right)=8x+12\)
\(\Leftrightarrow4x-4=8x+12\)
\(\Leftrightarrow-4x=16\)
\(\Leftrightarrow x=-4\left(L\right)\)
Vậy \(S=\varnothing\)
\(5\sqrt{x-1}-\sqrt{36\left(x-1\right)}+\sqrt{9\left(x-1\right)}=\sqrt{4\left(2x+3\right)}\)
\(5\sqrt{x-1}-6\sqrt{x-1}+3\sqrt{x-1}=2\sqrt{2x+3}\)
\(2\sqrt{x-1}=2\sqrt{2x+3}\)
\(\sqrt{x-1}=\sqrt{2x+3}\)
\(\hept{\begin{cases}2x+3\ge0\\x-1=2x-3\end{cases}}\)
\(\hept{\begin{cases}2x\ge-3\\x-2x=-3+1\end{cases}}\)
\(\hept{\begin{cases}x\ge-\frac{3}{2}\\-x=-2\end{cases}}\)
\(\hept{\begin{cases}x\ge-\frac{3}{2}\\x=2\end{cases}}\)
\(\Rightarrow x=2\)
Bạn FL.capheny bị sai dấu 2x+3 thành 2x-3 mà ai vẫn tk đúng vậy
Giải phương trình:
a)\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+2\sqrt{15}}\)
b)\(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
c) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
d) \(\sqrt{x^2-6x+9}+x=11\)
e) \(\sqrt{3x^2-4x+3}=1-2x\)
f) \(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
g) \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
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