(VD) Tìm x...
Đọc tiếp
(VD) Tìm x biết
a, format('truetype')%3Bfont-weight%3Anormal%3Bfont-style%3Anormal%3B%7D%3C%2Fstyle%3E%3C%2Fdefs%3E%3Cpolyline%20fill%3D%22none%22%20points%3D%2212%2C-18%2011%2C-18%205%2C0%202%2C-7%22%20stroke%3D%22%23000000%22%20stroke-linecap%3D%22square%22%20stroke-width%3D%221%22%20transform%3D%22translate(0.5%2C21.5)%22%2F%3E%3Cpolyline%20fill%3D%22none%22%20points%3D%225%2C0%202%2C-7%200%2C-6%22%20stroke%3D%22%23000000%22%20stroke-linecap%3D%22square%22%20stroke-width%3D%221%22%20transform%3D%22translate(0.5%2C21.5)%22%2F%3E%3Cline%20stroke%3D%22%23000000%22%20stroke-linecap%3D%22square%22%20stroke-width%3D%221%22%20x1%3D%2212.5%22%20x2%3D%2248.5%22%20y1%3D%223.5%22%20y2%3D%223.5%22%2F%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20font-style%3D%22italic%22%20text-anchor%3D%22middle%22%20x%3D%2218.5%22%20y%3D%2220%22%3Ex%3C%2Ftext%3E%3Ctext%20font-family%3D%22math135b31cfba37a56451b4768509d%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2230.5%22%20y%3D%2220%22%3E%26%23x2212%3B%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2242.5%22%20y%3D%2220%22%3E4%3C%2Ftext%3E%3Ctext%20font-family%3D%22math135b31cfba37a56451b4768509d%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2257.5%22%20y%3D%2220%22%3E%3D%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2275.5%22%20y%3D%2220%22%3E64%3C%2Ftext%3E%3C%2Fsvg%3E)
b, format('truetype')%3Bfont-weight%3Anormal%3Bfont-style%3Anormal%3B%7D%3C%2Fstyle%3E%3C%2Fdefs%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20font-style%3D%22italic%22%20text-anchor%3D%22middle%22%20x%3D%224.5%22%20y%3D%2218%22%3Ex%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2212%22%20text-anchor%3D%22middle%22%20x%3D%2212.5%22%20y%3D%2211%22%3E2%3C%2Ftext%3E%3Ctext%20font-family%3D%22math17f39f8317fbdb1988ef4c628eb%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2224.5%22%20y%3D%2218%22%3E%3D%3C%2Ftext%3E%3Ctext%20font-family%3D%22Arial%22%20font-size%3D%2216%22%20text-anchor%3D%22middle%22%20x%3D%2242.5%22%20y%3D%2218%22%3E64%3C%2Ftext%3E%3C%2Fsvg%3E)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
ET
5
6 tháng 12 2021
a) x-12=-15
x=-15+12
x=-3
b) 123-5(x+4)=38
5(x+4)=123-38
5(x+4)=85
x+4=85:5=17
x=17-4
x=13
DT
6 tháng 12 2021
\(a.x-12=-15\) \(b.123-5.\left(x+4\right)=38\)
\(x=\left(-15\right)+12\) \(5.\left(x+4\right)=123-38\)
\(x=-3\) \(5.\left(x+4\right)=85\)
\(x+4=85:5=17\)
\(x=17-4=13\)
T
10 tháng 8 2023
a) (x - 6)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)
b) \(\left|x\right|=3\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
N
1
DN
23 tháng 12 2021
a.\(2^x-2^4.2^7.32=0\)
\(2^x-2^{16}=0\)
\(=>x=16\)
b.\(3^x+3^{x+2}=270\)
\(3^x+3^x.3^2=270\)
\(3^x.10=270\)
\(3^x=27\)
\(=>x=3\)
NT
1
8 tháng 8 2021
a) \(0,\left(31\right)+x=0,3\left(7\right)\\ \Rightarrow\dfrac{31}{99}+x=\dfrac{17}{45}\\ \Rightarrow x=\dfrac{17}{45}-\dfrac{31}{99}=\dfrac{32}{495}=0,0\left(64\right)\)
Vậy \(x=0,0\left(64\right)\)
b) \(0,\left(4\right)\cdot x=\dfrac{5}{6}\\ \Rightarrow\dfrac{4}{9}\cdot x=\dfrac{5}{6}\\ \Rightarrow x=\dfrac{5}{6}:\dfrac{4}{9}\\ \Rightarrow x=\dfrac{5}{6}\cdot\dfrac{9}{4}\\ \Rightarrow x=\dfrac{15}{8}=1,875\)
Vậy \(x=1,875\)
KS
4 tháng 8 2023
`2x-15 = 17``
`=> 2x = 17 + 15`
`=> 2x = 32`
`=> X = 32 : 2`
`=> x = 16`
`156 - (x + 61) = 82`
`=> x + 61 = 156 - 82`
`=> x + 61 = 74`
`=> x = 13`
`2x - 138 = 2^3 . 3^2`
`=>2x - 138 = 72`
`=> 2x = 210`
`=> x = 105`
bài 2:
`23-3x = 8`
`=> 3x = 23 - 8`
`=> 3x = 15`
`=> x = 5`
`(x-35) - 120 = 0`
`=>(x-35) = 120`
`=> x = 120 +35`
`=> x = 155`
`3^x + 2 = 29`
`=> 3^x = 27`
`=> 3^x = 3^3`
`=> x = 3`
17 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`(2x - 1)^2 - 25 = 0`
`<=> (2x - 1)^2 = 25`
`<=> (2x - 1)^2 = (+-5)^2`
`<=>`\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy, `S = {-2; 3}`
`b,`
`8x^3 - 50x = 0`
`<=> x(8x^2 - 50) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2-50=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\8x^2=50\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x^2=\dfrac{25}{4}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=\pm\dfrac{5}{2}\end{matrix}\right.\)
Vậy, `S = {-5/2; 0; 5/2}.`
KV
17 tháng 7 2023
a) (2x - 1)² - 25 = 0
(2x - 1)² - 5² = 0
(2x - 1 - 5)(2x - 1 + 5) = 0
(2x - 6)(2x + 4) = 0
2x - 6 = 0 hoặc 2x + 4 = 0
*) 2x - 6 = 0
2x = 6
x = 3
*) 2x + 4 = 0
2x = -4
x = -2
Vậy x = -2; x = 3
b) 8x³ - 50x = 0
2x(4x² - 25) = 0
2x[(2x)² - 5²] = 0
2x(2x - 5)(2x + 5) = 0
2x = 0 hoặc 2x - 5 = 0 hoặc 2x + 5 = 0
*) 2x = 0
x = 0
*) 2x - 5 = 0
2x = 5
x = 5/2
*) 2x + 5 = 0
2x = -5
x = -5/2
Vậy x = -5/2; x = 0; x = 5/2
)3 = 
QN
2
TA
22 tháng 7 2021
a) `4\sqrt(2x-1)>8`
`<=>\sqrt(2x-1)>2`
`<=>2x-1>4`
`<=>x>5/2`
b) `2\sqrtx-1>3`
`<=>2\sqrtx>4`
`<=>\sqrtx>2`
`<=>x>4`
22 tháng 7 2021
a) Ta có: \(4\sqrt{2x-1}>8\)
\(\Leftrightarrow2x-1>4\)
\(\Leftrightarrow2x>5\)
hay \(x>\dfrac{5}{2}\)
b) Ta có: \(2\sqrt{x}-1>3\)
\(\Leftrightarrow\sqrt{x}>2\)
hay x>4
25 tháng 6 2021
a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-1\\x>1\end{matrix}\right.\)\(\Rightarrow x>1\)
Ta có : \(PT\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)
\(\Leftrightarrow x+1=4x-4\)
\(\Leftrightarrow3x=5\)
\(\Leftrightarrow x=\dfrac{5}{3}\left(TM\right)\)
Vậy ...
b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge1\\x>-1\end{matrix}\right.\)\(\Rightarrow x\ge1\)
Ta có : \(PT\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow x-1=4x+4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\left(L\right)\)
Vậy phương trình vô nghiệm .
25 tháng 6 2021
a) ĐKXĐ: \(x>1\)
Ta có: \(\dfrac{\sqrt{x+1}}{\sqrt{x-1}}=2\)
\(\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)
\(\Leftrightarrow x+1=4x-4\)
\(\Leftrightarrow x-4x=-4-1\)
\(\Leftrightarrow-3x=-5\)
hay \(x=\dfrac{5}{3}\left(nhận\right)\)
Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)
Ta có: \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}=2\)
\(\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)
\(\Leftrightarrow x-1=4x+4\)
\(\Leftrightarrow x-4x=4+1\)
\(\Leftrightarrow-3x=5\)
hay \(x=-\dfrac{5}{3}\)(loại)
Vậy: \(S=\varnothing\)
YP
1
21 tháng 11 2021
\(\left(-\dfrac{3}{4}x+1\right)\div\dfrac{2}{3}=1\)
\(-\dfrac{3}{4}x+1=1\times\dfrac{2}{3}\)
\(-\dfrac{3}{4}x+1=\dfrac{2}{3}\)
\(-\dfrac{3}{4}x=\dfrac{2}{3}-1\)
\(-\dfrac{3}{4}x=-\dfrac{1}{3}\)
\(x=-\dfrac{1}{3}\div\left(-\dfrac{3}{4}\right)\)
\(x=\dfrac{4}{9}\)
x+3=6
x=6-3
x=3
a) \(\sqrt{x-4}=64\left(ĐK:x\ge4\right)\)
\(\Rightarrow\left(\sqrt{x-4}\right)^2=64^2\)
\(\Rightarrow x-4=4096\)
\(\Rightarrow x=4100\)
b) \(x^2=64\)
\(\Rightarrow x^2=8^2\)
\(\Rightarrow x=8\)