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`@` ` \text {Ans}`

`\downarrow`

`a,`

`1/4+3/4*x=3/2-x`

`=> 1/4 + 3/4x - 3/2 + x = 0`

`=> (1/4 - 3/2) + (3/4x + x) = 0`

`=> -5/4 + 7/4x = 0`

`=> 7/4x = 5/4`

`=> x = 5/4 \div 7/4`

`=> x = 5/7`

Vậy, `x=5/7`

`b,`

`3/5*x-1/4=1/10*x-1/2`

`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`

`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`

`=> 1/2x + 1/4 = 0`

`=> 1/2x = -1/4`

`=> x = -1/4 \div 1/2`

`=> x = -1/2`

Vậy, `x=-1/2`

`c,`

`3x-3/5=x-1/4`

`=> 3x - 3/5 - x + 1/4 = 0`

`=> (3x - x) - (3/5 - 1/4) = 0`

`=> 2x - 7/20 = 0`

`=> 2x = 0,35`

`=> x = 0,35 \div 2`

`=> x = 7/40`

Vậy, `x=7/40`

`d,`

`3/2*x-2/5=1/3*x-1/4`

`=>  3/2x - 2/5 - 1/3x + 1/4 = 0`

`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`

`=> 7/6x - 3/20 = 0`

`=> 7/6x = 3/20`

`=> x = 3/20 \div 7/6`

`=> x = 9/70`

Vậy, `x=9/70`

`@` `\text {Kaizuu lv uuu}`

16 tháng 1 2022

a x=4

 

16 tháng 1 2022

\(a,\dfrac{x}{2}=\dfrac{8}{x}\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\\ b,\dfrac{x+1}{5}=\dfrac{x+1}{5}\left(luôn.đúng\right)\\ c,\dfrac{x+1}{5}=\dfrac{x+3}{10}\\ \Rightarrow\dfrac{2x+2}{10}=\dfrac{x+3}{10}\\ \Rightarrow2x+2=x+3\\ \Rightarrow2x-x=3-2\\ \Rightarrow x=1\\ d,\dfrac{x}{4}=\dfrac{18}{x+1}\\ \Rightarrow x\left(x+1\right)=4.18\\ \Rightarrow x^2+x=72\\ \Rightarrow x^2+x-72=0\\ \Rightarrow\left(x^2+9x\right)-\left(8x+72\right)=0\\ \Rightarrow x\left(x+9\right)-8\left(x+9\right)=0\\ \Rightarrow\left(x-8\right)\left(x+9\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-9\end{matrix}\right.\)

 

21 tháng 2 2023

` 8/23 . 46/24 =1/3 .x`

`=>8/23 . 23/12 =1/3 . x`

`=> 1/3 . x=2/3`

`=>x=2/3 : 1/3`

`=>x=2/3 . 3`

`=> x= 6/3`

`=>x=2`

`----`

`1/5 : x= 1/5-1/7`

`=>1/5 : x=  7/35 - 5/35`

`=> 1/5 :x= 2/35`

`=>x= 1/5 : 2/35`

`=>x=1/5 . 35/2`

`=>x=7/2`

`----`

`4/9 - (x-1/2)^2 =1/3`

`=> (x-1/2)^2 =4/9-1/3`

`=> (x-1/2)^2 =4/9- 3/9`

`=> (x-1/2)^2 =1/9`

`=> (x-1/2)^2 = (+- 1/3)^2`

`@ TH1`

`x-1/2=1/3`

`=>x=1/3+1/2`

`=>x= 2/6 + 3/6`

``=>x= 5/6`

`@ TH2`

`x-1/2=-1/3`

`=>x=-1/3 +1/2`

`=>x= -2/6 + 3/6`

`=>x=1/6`

`----`

`3,2 . x-(4/5+2/3) : 3 2/3 = 7/10`

`=> 3,2 . x-22/15 : 11/3 = 7/10`

`=>  3,2 . x-22/15 = 7/10 . 11/3`

`=>  3,2 . x-22/15 =77/30`

`=> 3,2 .x= 77/30 + 22/15`

`=> 3,2 .x=121/30`

`=>x= 121/30. 5/16`

`=>x= 121/96`

16 tháng 1

a) \(\dfrac{x}{4}\)=\(\dfrac{1}{x}\)→x2=4→x=2 hoặc x=-2
b) \(\dfrac{1}{5}\) = x:4-\(\dfrac{1}{10}\)→x:4=\(\dfrac{1}{5}\)+\(\dfrac{1}{10}\)→x:4=\(\dfrac{2+1}{10}\)→x:4=\(\dfrac{3}{10}\)→x=\(\dfrac{3}{10}\)x4→x=\(\dfrac{12}{10}\)→x=\(\dfrac{6}{5}\)

b: \(\Leftrightarrow\left(x-\dfrac{1}{2}\right):\dfrac{1}{3}=9+\dfrac{5}{7}-\dfrac{5}{7}=9\)

=>x-1/2=27

hay x=55/2

c: =>1/2x-3/4=42/63=2/3

=>1/2x=17/12

hay x=17/6

23 tháng 7 2021

b)(x+3)2-(x-4)(x+8)=1

\(\Rightarrow\)x2+6x+9-(x2+8x-4x-32)=1

⇒x2+6x+9-x2-8x+4x+32=1

⇒2x+41=1

\(\Rightarrow\)2x+41-1=0

\(\Rightarrow\)2x+40=0

⇒2x=-40

\(\Rightarrow\)x=\(\dfrac{-40}{2}\)

⇒x=-20

\(\text{#TNam}\)

`a,` Vì `y` tỉ lệ thuận với `x` theo hệ số tỉ lệ `k -> y=k*x`

Thay `x=4, y=3` vào ct

`-> 3=k*4`

`-> k=3/4`

Vậy, hệ số tỉ lệ `k=3/4`

`-> y=3/4 *x`

`b,` Khi `x=-12 -> y=3/4*(-12)=-9`

Khi `x=1/3 -> y=3/4*1/3=1/4`

Giải:

a) \(2^5=4^x\) 

\(\Rightarrow2^5=\left(2^2\right)^x\) 

\(\Rightarrow2^5=2^{2x}\) 

\(\Rightarrow2x=5\) 

\(\Rightarrow x=\dfrac{5}{2}\) 

b) \(2.4^2.8^3.16^4=8^x\) 

\(\Rightarrow2.\left(2^2\right)^2.\left(2^3\right)^3.\left(2^4\right)^4=\left(2^3\right)^x\) 

\(\Rightarrow2.2^4.2^9.2^{16}=2^{3x}\) 

\(\Rightarrow2^{30}=2^{3x}\) 

\(\Rightarrow3x=30\) 

\(\Rightarrow x=30:3\) 

\(\Rightarrow x=10\) 

c) \(3^3:3^5=9^x\) 

\(\Rightarrow3^{-2}=\left(3^2\right)^x\) 

\(\Rightarrow3^{-2}=3^{2x}\) 

\(\Rightarrow2x=-2\) 

\(\Rightarrow x=-2:2\)

\(\Rightarrow x=-1\) 

Chúc bạn học tốt!

a) Ta có: \(2^5=4^x\)

nên \(2^{2x}=2^5\)

\(\Leftrightarrow2x=5\)

hay \(x=\dfrac{5}{2}\)

b) Ta có: \(2\cdot4^2\cdot8^3\cdot16^4=8^x\)

\(\Leftrightarrow2^{3x}=2\cdot2^5\cdot2^9\cdot2^{16}=2^{31}\)

\(\Leftrightarrow3x=31\)

hay \(x=\dfrac{31}{3}\)

c) Ta có: \(3^3:3^5=9^x\)

\(\Leftrightarrow3^{-2}=3^{2x}\)

\(\Leftrightarrow2x=-2\)

hay x=-1

b) Ta có: \(B=x^2+2x+y^2-4y+6\)

\(=x^2+2x+1+y^2-4y+4+1\)

\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)

c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)

 

29 tháng 6 2021

\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)

\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)

\(B=x^2+2x+y^2-4y+6\)

\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)

\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2

\(C=4x^2+4x+9y^2-6y-5\)

\(=4x^2+4x+1+9y^2-6y+1-7\)

\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)

dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)

\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)

=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)

\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)

\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)

dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

20 tháng 12 2021

BS: \(x\in \mathbb{Z}\)

\(a,\Rightarrow x\in\left\{1;2;3;4\right\}\\ b,\Rightarrow x\in\left\{-3;-2;-1\right\}\\ d,\Rightarrow x\in\left\{-1;-2;-3;...\right\}\\ e,\Rightarrow x\in\left\{-4;-5;-6;...\right\}\)

20 tháng 12 2021

a.\(x\in\left\{1;2;3;4\right\}\)
b.\(x\in\left\{-3;-2;-1\right\}\)