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27 tháng 6 2015

a) \(\frac{x+4}{20}=\frac{5}{x+4}\Leftrightarrow\left(x+4\right)^2=100\Rightarrow x+4=+-10\Leftrightarrow x=-4+-10\)

b) \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\Leftrightarrow x^2+2x-3=x^2-4\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)

 

27 tháng 6 2015

a)\(\frac{x+4}{20}=\frac{5}{x+4}\)

<=>(x+4)2=20.5

<=>(x+4)2-100=0

<=>(x+4-10)(x+4+10)=0

<=>(x-6)(x+14)=0

<=>x-6=0 hoặc x+14=0

<=>x=6 hoặc x=-14

b)\(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)

<=>(x-1)(x+3)=(x-2)(x+2)

<=>x2+2x-3=x2-4

<=>x2+2x-3-x2+4=0

<=>2x+1=0

<=>2x=-1

<=>x=-1/2

31 tháng 1 2022

Bài 4:

a) \(\dfrac{x}{2}=\dfrac{2}{4}\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{1}{2}\)

\(\Rightarrow x=2\)

Vậy: \(x=2\)

b) \(-\dfrac{1}{5}=\dfrac{2}{x}\)

\(\Rightarrow x=\dfrac{-5.2}{1}=-10\)

Vậy: \(x=-10\)

c) \(\dfrac{x}{5}=\dfrac{5}{x}\)

\(\Rightarrow x^2=25\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{5;-5\right\}\)

31 tháng 1 2022

\(\dfrac{x}{2}=\dfrac{2}{4}\\ =>x=\dfrac{2.2}{4}=1\)

\(\dfrac{-1}{5}=\dfrac{2}{x}\\ =>x=\dfrac{2.5}{-1}=-10\)

\(\dfrac{x}{5}=\dfrac{5}{x}\\ =>x^2=25\\ x=5;x=-5\)

2 tháng 5 2021

a, \(\left(\dfrac{1}{2}+\dfrac{4}{7}\right):x=\dfrac{-3}{4}\)

\(\dfrac{15}{14}:x=\dfrac{-3}{4}\)

=> x= \(\dfrac{-7}{10}\)

b, 0,5:x-\(1\dfrac{3}{4}\)= 25%

0,5:x-\(\dfrac{7}{4}=\dfrac{1}{4}\)

0,5:x = 2

=> x = \(\dfrac{1}{4}\)

28 tháng 6 2021

`|2x+1|-3=x+4`

`<=>|2x+1|=x+4+3=x+7(x>=-7)`

`**2x+1=x+7`

`<=>x=7-1=6(tm)`

`**2x+1=-x-7`

`<=>3x=-6`

`<=>x=-2(tm)`

`|3x-5|=1-3x(x<=1/3)`

`**3x-5=1-3x`

`<=>6x=6`

`<=>x=1(l)`

`**3x-5=3x-1`

`<=>-5=-1` vô lý

`|2x+2|+|x-1|=10`

Nếu `x>=1`

`pt<=>2x+2+x-1=10`

`<=>3x+1=10`

`<=>3x=9`

`<=>x=3(tm)`

Nếu `x<=-1`

`pt<=>-2x-2+1-x=10`

`<=>-1-3x=10`

`<=>-11=3x`

`<=>x=-11/3(tm)`

Nếu `-1<=x<=1`

`pt<=>2x+2+1-x=10`

`<=>x+3=10`

`<=>x=7(l)`

Vậy `S={3,-11/3}`

pt là phương trình phải ko vậy?

 

a: \(\Leftrightarrow2\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}=28\)

=>\(13\sqrt{2x}=28\)

=>căn 2x=28/13

=>2x=784/169

=>x=392/169

b: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>2*căn x-5=4

=>căn x-5=2

=>x-5=4

=>x=9

c: =>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=>x-2=0 hoặc x+2=1

=>x=-1 hoặc x=2

4 tháng 3 2022

\(a,\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{5}{6}-\dfrac{2}{3}\\ \Rightarrow\dfrac{3}{4}x-\dfrac{7}{12}=\dfrac{1}{6}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{1}{6}+\dfrac{7}{12}\\ \Rightarrow\dfrac{3}{4}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{3}{4}:\dfrac{3}{4}\\ \Rightarrow x=1\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{5}{7}\\ \Rightarrow\dfrac{-5}{x}=\dfrac{-5}{-7}\\ \Rightarrow x=-7\\ c,2\dfrac{1}{3}:x=7\\ \Rightarrow\dfrac{7}{3}:x=7\\ \Rightarrow x=\dfrac{7}{3}:7\\ \Rightarrow x=\dfrac{1}{3}\)

\(d,\dfrac{-105}{12}< x< \dfrac{20}{7}\Rightarrow x\in\left\{-8;-7;...;2\right\}\)

a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=\dfrac{3}{4}\)

hay x=1

b: \(\Leftrightarrow x=\dfrac{-28\cdot5}{20}=-7\)

c: \(\Leftrightarrow x=\dfrac{7}{3}:7=\dfrac{1}{3}\)

d: \(\Leftrightarrow-8< x< 3\)

hay \(x\in\left\{-7;-6;-5;-4;-3;-2;-1;0;1;2\right\}\)

26 tháng 6 2021

`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}`

AH
Akai Haruma
Giáo viên
26 tháng 6 2021

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 $a=4$

$\Leftrightarrow \sqrt{2x-1}=4$

$\Leftrightarrow x=\frac{17}{2}$

27 tháng 9 2021

a) \(\Leftrightarrow2\left|3x-1\right|=\dfrac{4}{5}\)

\(\Leftrightarrow\left|3x-1\right|=\dfrac{2}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=\dfrac{2}{5}\\3x-1=-\dfrac{2}{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{15}\\x=\dfrac{1}{5}\end{matrix}\right.\)

b)TH1:  \(x\ge3\)

\(\Leftrightarrow x+5+x-3=9\Leftrightarrow2x=7\Leftrightarrow x=\dfrac{7}{2}\left(tm\right)\)

TH2: \(-5\le x< 3\)

\(\Leftrightarrow x+5-x+3=9\Leftrightarrow8=9\left(VLý\right)\)

TH3: \(x< -5\)

\(\Leftrightarrow-x-5-x+3=9\Leftrightarrow2x=-11\Leftrightarrow x=-\dfrac{11}{2}\left(tm\right)\)

27 tháng 9 2021

\(a,2.|3x-1|-\dfrac{3}{4}=\dfrac{1}{20}\)

\(2.|3x-1|=\dfrac{1}{20}+\dfrac{3}{4}\)

\(2.|3x-1|=\dfrac{4}{5}\)

\(|3x-1|=\dfrac{4}{5}:2\)

\(|3x-1|=\dfrac{2}{5}\)

\(\Rightarrow3x-1=\pm\dfrac{2}{5}\)

\(3x-1=\dfrac{2}{5}\)

\(3x=\dfrac{2}{5}+1\)

\(3x=\dfrac{7}{5}\)

\(x=\dfrac{7}{5}:3\)

\(x=\dfrac{7}{15}\)

\(3x-1=-\dfrac{2}{5}\)

\(3x=-\dfrac{2}{5}+1\)

\(3x=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:3\)

\(x=\dfrac{1}{5}\)

`@` ` \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}`