GIÚP MIK VỚI
Tìm x:
a, |x - 2| + |x+ 7,5| = 9,5
b, 2|x+3 | + |2x + 5| = 11
3. Tìm x:a
)2+x=13/2.
b)x:3/4=4/5 x=......
Giúp mik vs nha mik cần gấp ạ 🥺🥺🥺
\(a,x=\dfrac{13}{2}-2\\ x=\dfrac{9}{2}\\ b,x=\dfrac{4}{5}\times\dfrac{3}{4}\\ x=\dfrac{12}{20}=\dfrac{3}{5}\)
tìm x:
a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)
b, 2(x - 1) - 5 (x + 2) = -10
mong các bạn giúp mình :333
Arigato :333
a) Ta có: \(\left(2x-3\right)-\left(x-5\right)=\left(x+2\right)-\left(x-1\right)\)
\(\Leftrightarrow2x-3-x+5=x+2-x+1\)
\(\Leftrightarrow x+2=3\)
hay x=1
Vậy: x=1
b) Ta có: \(2\left(x-1\right)-5\left(x+2\right)=-10\)
\(\Leftrightarrow2x-2-5x-10=-10\)
\(\Leftrightarrow-3x=-10+10+2=2\)
hay \(x=-\dfrac{2}{3}\)
Vậy: \(x=-\dfrac{2}{3}\)
a, (2x - 3) - (x - 5) = (x + 2) - (x - 1)
2x - 3 - x + 5 = x + 2 - x + 1
(2x - x) + (-3 + 5) = (x - x) + (2 + 1)
x + 2 = 3
x = 1
Mọi người ơi giúp mình bài này với ạ,mai mình học rồi , cho mình câu trả lời rõ ý nhất với nha, mik cảm ơn ạ.
bài tìm x:
a,2x2+3(x-1)(x+1)=5x(x+1)
b,(x-3)3+3-x=0
c,5x(x-2000)-x+2000=0
d, 3(2x - 3) + 2(2 - x) = -3
e,x+6x2=0
`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
__________________________________________
`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
__________________________________________
`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
__________________________________________
`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
__________________________________________
`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
Tìm x:
a) 1 = (2x + 0,5)600
b) (x - 0,125)2 = 0,25
c) (x - 3)11 = (x - 3)41
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
a, 1=(2x+0,5)600
=>1600=(2x+0,5)600
=>1=2x+0,5
=>2x+0,5=1
=> 2x= 1+0,5
=>2x= 1,5
=:>x= 1,5:2
=>x=0,75
b, (x-0,125)2=0,25
=>(x-0,125)2=(0,5)2
=>x-0,125=0,5
=>x=0,5 +0,125
=>x= 0,625
c, (x-3)11=(x-3)41
(x-3)11 -(x-3)41=0
(x-3)11-(x-3)11 .(x-3)30=0
(x-3)11.[1-(x-3)30 ]=0
(x-3)11.(4-x)30=0
(x-3)11=0 hoặc (4-x)30=0
(x-3)11=0 ( 4-x)30=0
x-3=0 4-x=0
x=0+3 x=4-0
x=3 x=4
vậy xϵ{3,4}
Tìm X:
a) (x + 3)2 – x2 + 15 = 1
b) (5 – x)2 + 6x – x2 = –7
c) (1 + x)(1 – x) – (x – 4)2 = –15
giúp mik với mik cần gấp!!
a)(x+3)2-x2+15=2x+6-2x+15=1
=21=1
Bạn chép sai đầu bài à
Tìm x:
a. x + 3/7 = 2/5 : 18/35
b. x nhân 5/9 = 4/5 - 1/3
Giải thích rõ ràng giúp mik nhé😢
a. x + \(\dfrac{3}{7}\)= \(\dfrac{2}{5}:\dfrac{18}{25}=>x+\dfrac{3}{7}=\dfrac{2}{5}\)x\(\dfrac{35}{18}=>x+\dfrac{3}{7}=\dfrac{7}{9}\)
=> x = \(\dfrac{7}{9}-\dfrac{3}{7}=\dfrac{49}{63}-\dfrac{27}{63}=\dfrac{22}{63}\)
b. \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{4}{5}-\dfrac{1}{3}\)
=> \(x\) x \(\dfrac{5}{9}\)= \(\dfrac{12}{15}-\dfrac{5}{15}=>x\) x \(\dfrac{5}{9}\)= \(\dfrac{7}{15}\)
=> x = \(\dfrac{7}{15}:\dfrac{5}{9}\)
=> x = \(\dfrac{21}{25}\)
a,x + 3/7 = 2/5 : 18/35
x + 3/7 = 7/9
x = 7/9 - 3/7
x = 22/63
vậy x = ...
b, X x 5/9 = 4/5 - 1/3
X x 5/9 = 7/15
X = 7/15 : 5/9
X = 21/25
vậy X = ...
Tìm x:
a. x + 3/7 = 2/5 : 18/35
b. x nhân 5/9 = 4/5 - 1/3
Giải thích rõ ràng giúp mik nhé😢
\(a.x+\dfrac{3}{7}=\dfrac{2}{5}:\dfrac{18}{35}\\x+\dfrac{3}{7}=\dfrac{2}{5}\times\dfrac{35}{18} \\ x+\dfrac{3}{7}=\dfrac{7}{9}\\ x=\dfrac{7}{9}-\dfrac{3}{7}\\ x=\dfrac{22}{63}\)
\(b.x\times\dfrac{5}{9}=\dfrac{4}{5}-\dfrac{1}{3}\\x\times\dfrac{5}{9}=\dfrac{7}{15}\\ x=\dfrac{7}{15}:\dfrac{5}{9}\\ x= \dfrac{21}{25}\)
mk đã giải cho bạn rồi nhé! bạn xem lại phần thông báo ạ!
Tìm x:
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\) b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
mọi người ơi giúp mik với ai làm đc mik tick cho
a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
⇔\(7\left(x-3\right)=5\left(x+5\right)\)
⇔\(7x-21=5x+25\)
⇔\(7x-21-5x-25=0\)
⇔\(2x-46=0\)
⇔\(2x=46\)
⇔\(x=23\)
b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)
⇔\(\left(x+1\right)\left(x-1\right)=7.9\)
⇔\(x^2-1=63\)
⇔\(x^2=64=8^2\)
⇔\(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
\(a.\dfrac{x-3}{x+5}=\dfrac{5}{7}\)
\(\left(x-3\right).7=\left(x+5\right).5\)
\(\left(x.7\right)+\left[\left(-3\right).7\right]=\left(x.5\right)+\left(5.5\right)\)
\(7x-21=5x+25\)
\(7x-5x=25+21\)
\(2x=46\)
\(x=46:2\)
\(x=23\)
Câu b/ cứ làm theo câu a/ là được
AcCl3 và NiCl2
Tìm x:
a) 4.(2-x)+x.(x+6)=x2
b) x.(x-7)-(x-2).(x+5)=0
c) (2x+3).(3-2x)+(2x-1)2=2
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2