Tìm x biết:
a, \(\sqrt{x-1}\) = 3 b,\(x^2\) - 64 = 0
c,\(x^2\) + 16 = 25 d,|\(\sqrt{x}-3\)| + 3 = 9
bài 3: Tìm x, biết:
a)(2x-1)2-25=0 b)8x3-50x=0
c)2(x+3)-x2-3x=0 d)x3+27+(x+3)(x-9)=0
mình cần gấppppppppp
bài 3: Tìm x, biết:
a)(2x-1)2-25=0 b)8x3-50x=0
c)2(x+3)-x2-3x=0 d)x3+27+(x+3)(x-9)=0
mình cần gấpppppppppp
a: Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
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)
Tìm x biết:
a)\(\sqrt{9x^2}=6\)
b)\(\sqrt{\left(x-2\right)^2}=5\)
c)\(\sqrt{x^2-6x+9}=3\)
d)\(\sqrt{x^2+4x+4}-2x=3\)
`a)sqrt{9x^2}=6`
`<=>|3x|=6`
`<=>|x|=2`
`<=>x=+-2`
`b)sqrt{(x-2)^2}=5`
`<=>|x-2|=5`
`**x-2=5`
`<=>x=7`
`**x-2=-5`
`<=>x=-3`
`c)sqrt{x^2-6x+9}=3`
`<=>\sqrt{(x-3)^2}=3`
`<=>|x-3|=3`
`**x-3=3`
`<=>x=6`
`**x-3=-3`
`<=>x=0`
`d)sqrt{x^2+4x+4}-2x=3`
`<=>sqrt{(x+2)^2}=3+2x`
`<=>|x+2|=2x+3(x>=-3/2)`
`**x+2=2x+3`
`<=>x=-1(tm)`
`**x+2=-2x-3`
`<=>3x=-5`
`<=>x=-5/3(l)`
Sử dụng công thức:`sqrtA^2=|A|`
ĐKXĐ : \(x\in R\)
a, \(\sqrt{9x^2}=\left|3x\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy ..
b, \(\sqrt{\left(x-2\right)^2}=\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy ...
c, \(\sqrt{x^2-6x+9}=\sqrt{\left(x-3\right)^2}=\left|x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=3\\x-3=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)
Vậy ..
d, \(\sqrt{x^2+4x+4}-2x=\sqrt{\left(x+2\right)^2}-2x=\left|x+2\right|-2x=3\)
\(\Leftrightarrow\left|x+2\right|=2x+3\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x+2=2x+3\\x+2=-2x-3\end{matrix}\right.\\2x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{3}{2}\\\left[{}\begin{matrix}x=-1\left(TM\right)\\x=-\dfrac{5}{3}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy ..
2) Tính: (Giải chi tiết từng bước)
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
3) Tìm x, biết:
a) \(\sqrt{\left(x-1\right)^2}=4\)
b) \(\sqrt{36x^2-60x+25}=4\)
Bài 2:
a) \(2\sqrt{125}+\dfrac{3}{2}\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\)
\(=2\sqrt{5^2\cdot5}+\dfrac{3}{2}\sqrt{4^2\cdot5}-\sqrt{6^2\cdot5}-\dfrac{2}{7}\sqrt{7^2\cdot5}\)
\(=10\sqrt{5}+\dfrac{3\cdot4}{2}\sqrt{5}-6\sqrt{5}-\dfrac{2\cdot7}{7}\sqrt{5}\)
\(=10\sqrt{5}+6\sqrt{6}-6\sqrt{5}-2\sqrt{5}\)
\(=8\sqrt{5}\)
b) \(\sqrt{11-4\sqrt{7}}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\cdot2\cdot\sqrt{7}+2^2}-\sqrt{\left(\sqrt{7}\right)^2+2\cdot3\cdot\sqrt{7}+3^2}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{\left(\sqrt{7}+3\right)^2}\)
\(=\sqrt{7}-2-\sqrt{7}-3\)
\(=-5\)
\(2a,\\ 2\sqrt{125}+\dfrac{3}{2}.\sqrt{80}-\sqrt{180}-\dfrac{2}{7}\sqrt{245}\\ =2\sqrt{5^2.5}+\dfrac{3}{2}.\sqrt{4^2.5}-\sqrt{6^2.5}-\dfrac{2}{7}.\sqrt{7^2.5}\\ =2.5.\sqrt{5}+\dfrac{3}{2}.4.\sqrt{5}-6\sqrt{5}-\dfrac{2}{7}.7\sqrt{5}\\ =10\sqrt{5}+6\sqrt{5}-6\sqrt{5}-2\sqrt{5}=8\sqrt{5}\)
3:
a: =>|x-1|=4
=>x-1=4 hoặc x-1=-4
=>x=-3 hoặc x=5
b: =>|6x-5|=4
=>6x-5=4 hoặc 6x-5=-4
=>6x=1 hoặc 6x=9
=>x=1/6 hoặc x=3/2
Câu 2: Tìm x biết:
a. \(\sqrt{x-3}=5\)
b. \(\sqrt{2x-1}=\sqrt{3}\)
c. \(\sqrt{1-x}=-1\)
d. \(\sqrt{\left(x-1\right)^2}=1\)
\(a,ĐK:x\ge3\\ PT\Leftrightarrow x-3=5\Leftrightarrow x=8\left(tm\right)\\ b,ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow2x-1=3\Leftrightarrow x=2\left(tm\right)\\ c,Vì.\sqrt{1-x}\ge0>-1.nên.pt.vô.nghiệm\\ d,PT\Leftrightarrow\left|x-1\right|=1\Leftrightarrow\left[{}\begin{matrix}x-1=1\\1-x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
a) \(\sqrt{x-3}=5\) (1)
ĐKXĐ: \(x\ge3\)
\(\left(1\right)\Leftrightarrow x-3=25\)
\(\Leftrightarrow x=28\) (nhận)
Vậy \(x=28\)
b) \(\sqrt{2x-1}=\sqrt{3}\) (2)
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(\left(2\right)\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\) (nhận)
Vậy \(x=2\)
c) \(\sqrt{1-x}=-1\)
Không tìm được \(x\) vì \(\sqrt{1-x}\ge0\) (với mọi \(x\le1\))
d) \(\sqrt{\left(x-1\right)^2}=1\) (3)
ĐKXĐ: Với mọi \(x\in R\)
\(\left(3\right)\Leftrightarrow\left|x-1\right|=1\)
\(\Leftrightarrow x-1=1\) (khi \(x\ge1\)) hoặc \(1-x=1\) (khi \(x< 1\))
* \(x-1=1\)
\(\Leftrightarrow x=2\) (nhận)
* \(1-x=1\)
\(\Leftrightarrow x=0\) (nhận)
Vậy \(x=0;x=2\)
Tìm x, biết:
a) \(\sqrt{\left(x-3\right)^2}=3-x\)
b) \(\sqrt{25-20x+4x^2}+2x=5\)
a,\(Đkxđ:x\ge3\)
Ta có:
\(\sqrt{\left(x-3\right)^2}=3-x\)
\(\Leftrightarrow|x-3|=3-x\)
\(\Leftrightarrow x-3=\left[{}\begin{matrix}x-3\\3-x\end{matrix}\right.\)
\(TH1:x-3=x-3\Leftrightarrow0x=0\)
\(\Rightarrow\)\(x\in R\) và \(x\ge3\)
\(TH2:x-3=3-x\Leftrightarrow2x=6\Leftrightarrow x=3\)( ko thỏa mãn điều kiện)
vậy \(\left\{x\in R/x\ge3\right\}\)
b, \(Đkxđ:x\le\dfrac{5}{2}\)
Ta có:
\(\sqrt{25-20x+4x^2}+2x=5\)
\(\Leftrightarrow\sqrt{\left(5-2x\right)^2}+2x=5\)
\(\Leftrightarrow\left|5-2x\right|=5-2x\)
\(\Leftrightarrow\left[{}\begin{matrix}5-2x=5-2x\\5-2x=2x-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in R\\x=\dfrac{5}{2}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(\left\{x\in R/x\le\dfrac{5}{2}\right\}\)
Câu 2: Tìm x biết:
a. \(\sqrt{\left(2x-3\right)^2}=7\)
b. \(\sqrt{64x-121}-\sqrt{25x-50}-\sqrt{4x-1}=20\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\)
a: \(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)