Tìm x biết a) 3√x - 7 = 0 b) √x-2 + √4x-8=3
Tìm x biết a) 3√x - 7 = 0 b) √x-2 + √4x-8=3
\(a,3\sqrt{x}-7=0\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{x}=7\\ \Leftrightarrow\sqrt{x}=\dfrac{7}{3}\\ \Leftrightarrow x=\dfrac{49}{9}\left(tmdk\right)\)
Vậy \(S=\left\{\dfrac{49}{9}\right\}\)
\(b,\sqrt{x-2}+\sqrt{4x-8}=3\left(dk:x\ge2\right)\\ \Leftrightarrow\sqrt{x-2}+\sqrt{4\left(x-2\right)}=3\\ \Leftrightarrow\sqrt{x-2}+2\sqrt{x-2}=3\\ \Leftrightarrow3\sqrt{x-2}=3\\ \Leftrightarrow\sqrt{x-2}=1\\ \Leftrightarrow x-2=1\\ \Leftrightarrow x=3\left(tmdk\right)\)
Vậy \(S=\left\{3\right\}\)
a: =>3*căn x=7
=>căn x=7/3
=>x=49/9
b: =>3*căn x-2=3
=>căn x-2=1
=>x-2=1
=>x=3
`a, 3 sqrt x - 7 = 0`
`<=> 3 sqrt x = 7`
`<=> sqrt x = 7/3`
`<=> x = 49/9`.
Vậy `x = 49/9`
`b, sqrt(x-2) + sqrt(4x-8) = 3`
`<=> sqrt(x-2) + 2 sqrt(x-2) = 3`
`<=> 3 sqrt(x-2) = 3`
`<=> sqrt(x-2)=1`
`<=> x-2=1`
`<=> x = 3`
Vậy `x = 3`
sqrt(x + 3) + sqrt(5 - x) + 6sqrt((x + 3)(5 - x)) = 2
`\sqrt{x+3}+\sqrt{5-x}+6\sqrt{(x+3)(5-x)}=2`
`->` đề như này à bạn
với x lớn hơn hoặc bằng 2, tìm giá trị ngộ nhất của 2cănx+1
giúp em với ạ
`1)\sqrt{3x}=6<=>3x=36<=>x=12`
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`2)\sqrt{5x}=\sqrt{10}<=>5x=10<=>x=2`
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`3)\sqrt{4-5x}=12<=>4-5x=144<=>x=-28`
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`4)\sqrt{1-4x+4x^2}=5`
`<=>|1-2x|=5`
`<=>[(1-2x=5),(1-2x=-5):}<=>[(x=-2),(x=3):}`
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`5)\sqrt{4x-20}+\sqrt{x-5}-1/3\sqrt{9x-45}=4` `ĐK: x >= 5`
`<=>2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4`
`<=>\sqrt{x-5}=2`
`<=>x-5=4`
`<=>x=9` (t/m)
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`6)1/2\sqrt{x-1}-3/2\sqrt{9x-9}+24\sqrt{[x-1]/64}=-17` `ĐK: x >= 1`
`<=>1/2\sqrt{x-1}-9/2\sqrt{x-1}+3\sqrt{x-1}=-17`
`<=>\sqrt{x-1}=17`
`<=>x-1=289`
`<=>x=290` (t/m)
`p = ((a + 2 sqrt a + 1)/(sqrt a + 1)) ((a - 2sqrt a + 1)/(sqrt a-1))`
`= (sqrt a + 1)(sqrt a - 1)`
`= a - 1`
tìm x để một căn thức sau luôn có nghĩa
a)√3-x
b)√-4+16
c)√-3/2x+10
d)√25/-4+12
e)√5x^2+1
f)√1/3x^2-5
g)√x^2-6x+9
a: ĐKXĐ: 3-x>=0
hay x<=3
c: ĐKXĐ: -3/2x+10>=0
=>-3/2x>=-10
=>3/2x<=10
=>x<=10:3/2=20/3
e: ĐKXĐ: \(x\in R\)
f: ĐKXĐ: \(3x^2-5>0\)
hay \(x\in\left(-\infty;-\sqrt{\dfrac{5}{3}}\right)\cup\left(\sqrt{\dfrac{5}{3}};+\infty\right)\)
Giúp với em đang gấp
a, \(\Delta=\left(-3\right)^2-4.1.\left(-5\right)=9+20=29>0\)
Suy ra pt luôn có 2 nghiệm phân biệt
Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=3\\x_1x_2=-5\end{matrix}\right.\)
\(x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=3^2-2.\left(-5\right)=9+10=19\)
\(b,A=\dfrac{x_1-2}{x_2}+\dfrac{x_2-2}{x_1}\\ =\dfrac{x_1\left(x_1-2\right)+x_2\left(x_2-2\right)}{x_1x_2}\\ =\dfrac{x_1^2-2x_1+x_2^2-2x_2}{-5}\\ =\dfrac{\left(x_1+x_2\right)^2-2\left(x_1+x_2\right)}{-5}\\ =\dfrac{3^2-2.3}{-5}\\ =\dfrac{-3}{5}\)
giải pt:
\(2x^2+1=\dfrac{1}{x^2}-4\)
cần gấp ạ
Đặt x2 = t > 0 ta được
\(2t+1=\dfrac{1}{t}-4\Leftrightarrow2t^2+5t-1=0\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{33}}{4}\\t=\dfrac{-5-\sqrt{33}}{4}\left(loại\right)\end{matrix}\right.\\ \Leftrightarrow x^2=\dfrac{-5+\sqrt{33}}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\sqrt{-5+\sqrt{33}}}{2}\\x=\dfrac{\sqrt{-5+\sqrt{33}}}{2}\end{matrix}\right.\)
Vậy pt có 2 nghiệm
\(2x^2+1=\dfrac{1}{x^2}-4\left(1\right)\)
Đặt \(x^2=t\left(t\ge0\right)\)
Khi đó phương trình \(\left(1\right)\) trở thành \(2t+1=\dfrac{1}{t}-4\)
\(\Leftrightarrow2t^2+5t-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-5+\sqrt{33}}{4}\left(\text{nhận}\right)\\t=\dfrac{-5-\sqrt{33}}{4}\left(\text{loại}\right)\end{matrix}\right.\)
\(\Rightarrow x^2=\dfrac{-5+\sqrt{33}}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\sqrt{-5+\sqrt{33}}}{2}\\x=\dfrac{\sqrt{-5+\sqrt{33}}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{-\sqrt{-5+\sqrt{33}}}{2};\dfrac{\sqrt{-5+\sqrt{33}}}{2}\right\}\)
b) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(\Leftrightarrow4\sqrt{3x-2}+2\sqrt{2x-1}=2\sqrt{3x-2}+4\sqrt{2x-1}\\ \Leftrightarrow2\sqrt{3x-2}=2\sqrt{2x-1}\\ \Leftrightarrow\sqrt{3x-2}=\sqrt{2x-1}\\ \Leftrightarrow3x-2=2x-1\\ \Leftrightarrow x=1\)
c) ĐKXĐ:\(x\ge\dfrac{2}{3}\)
\(\Leftrightarrow5\sqrt{3x-2}+3\sqrt{2x-1}=4\sqrt{3x-2}+4\sqrt{2x-1}\\ \Leftrightarrow\sqrt{3x-2}=\sqrt{2x-1}\\ \Leftrightarrow3x-2=2x-1\\ \Rightarrow x=1\)
\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow4\sqrt{3x-2}+2\sqrt{2x-1}=2\sqrt{3x-2}+4\sqrt{2x-1}\\ \Leftrightarrow2\sqrt{3x-2}=2\sqrt{2x-1}\\ \Leftrightarrow3x-2=2x-1\\ \Leftrightarrow x=1\left(tm\right)\\ c,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow5\sqrt{3x-2}+3\sqrt{2x-1}=4\sqrt{3x-2}+4\sqrt{2x-1}\\ \Leftrightarrow\sqrt{3x-2}=\sqrt{2x-1}\\ \Leftrightarrow3x-2=2x-1\Leftrightarrow x=1\left(tm\right)\)
b) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow2\sqrt{12x-8}+2\sqrt{2x-1}=\sqrt{12x-8}+4\sqrt{2x-1}\)
\(\Leftrightarrow\sqrt{12x-8}-2\sqrt{2x-1}=0\)
\(\Leftrightarrow12x-8+8x-4=2.2\sqrt{\left(12x-8\right)\left(2x-1\right)}\)
\(\Leftrightarrow5x-3=\sqrt{24x^2-28x+8}\)
\(\Leftrightarrow25x^2-30x+9=24x^2-28x+8\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\left(tm\right)\)
c) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow5\sqrt{3x-2}+3\sqrt{2x-1}=4\sqrt{3x-2}+4\sqrt{2x-1}\)
\(\Leftrightarrow\sqrt{3x-2}=\sqrt{2x-1}=\)
\(\Leftrightarrow3x-2=2x-1\Leftrightarrow x=1\left(tm\right)\)
Bài 2:
Gọi số học sinh nam là x
Số học sinh nữ là y
Theo đề, ta có: \(\left\{{}\begin{matrix}x+y=40\\\dfrac{5}{7}x+\dfrac{3}{4}y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{7}x+\dfrac{5}{7}y=\dfrac{200}{7}\\\dfrac{5}{7}x+\dfrac{3}{4}y=11\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{1}{28}y=\dfrac{123}{y}\\\end{matrix}\right.\)
=> Đề sai rồi bạn