a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{8}{3}=2\)

=>2x=-3/2

hay x=-3/4

b: 2x+3=5

=>2x=2

hay x=1

c: =>3(x-2)=4(5+x)

=>4x+20=3x-6

=>x=-26

a) => (7/2 + 2x) . 8/3 = 16/3
=> 7/2 + 2x = 16/3 : 8/3
=> 7/2 + 2x = 2
=> 2x = 2 - 7/2
=> 2x = -1.5
=> x = -1.5 : 2
=> x = -0.1

a: \(\Leftrightarrow2x+\dfrac{7}{2}=\dfrac{16}{3}:\dfrac{11}{3}=\dfrac{16}{11}\)

=>2x=-45/22

hay x=-45/44

b: =>x/7=-1/28:1/4=-1/7

=>x=-1

a)(7/2+2x).11/3=16/3

7/2+2x=16/3:11/3

7/2+2x=16/3.3/11

7/2+2x=16/11

2x=16/11-7/2

2x= -45/22

x= -45/22:2

x= -45/44

Vậy x= -45/44

b)x/7 +1/4= -1/28

x/7= -1/28-1/4

x/7= -2/7

=>x= -2

a/  => \(\dfrac{3}{5}.\dfrac{1}{x}=\dfrac{6}{25}\)

=> \(\dfrac{1}{x}=\dfrac{2}{5}\)

=> x = 5/2

b/ \(\Rightarrow2\left(x-\dfrac{1}{3}\right)=\dfrac{2}{15}\)

=> \(x-\dfrac{1}{3}=\dfrac{1}{15}\)

=> \(x=\dfrac{2}{5}\)

c/ => | x + 1| = 10/21

=> \(\left[{}\begin{matrix}x=-\dfrac{11}{21}\\x=-\dfrac{31}{21}\end{matrix}\right.\)

d/ => \(5x+5=6x-3\)

=> x = 8

a) (x-1):2/3=-2/5

=>x-1=-4/15

=>x=11/15

b) |x-1/2|-1/3=0

=>|x-1/2|=1/3

=>\(\left\{{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\) 

c) Tương Tự câu B

a, Ta có : 

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\Rightarrow\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

\(\Rightarrow x=11;y=17;z=23\)

b, Đặt \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\Rightarrow xyz=810\)

\(\Rightarrow2k.3k.5k=810\Leftrightarrow30k^3=810\Leftrightarrow k^3=27\Leftrightarrow k=3\)

\(\Rightarrow x=6;y=9;z=15\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{2x-2}{4};\dfrac{y-2}{3}=\dfrac{3y-6}{9};\dfrac{z-3}{4}\)

Áp dụng t/c dtsbn:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x-2+3y-6-z+3}{4+9-4}=5\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=5\\\dfrac{y-2}{3}=5\\\dfrac{z-3}{4}=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=12\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

xyz = 810

=> 2k.3k.5k = 810

=> k = 3

\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=9\\z=15\end{matrix}\right.\)

a) Ta có: \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

nên \(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}\)

mà 2x+3y-z=50

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{2x-2}{4}=\dfrac{3y-6}{9}=\dfrac{z-3}{4}=\dfrac{2x+3y-z-2-6+3}{4+9-4}=\dfrac{50-5}{9}=5\)

Do đó:

\(\left\{{}\begin{matrix}x-1=10\\y-2=15\\z-3=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=11\\y=17\\z=23\end{matrix}\right.\)

b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Ta có: xyz=810

\(\Leftrightarrow30k^3=810\)

\(\Leftrightarrow k^3=27\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=2\cdot3=6\\y=3k=3\cdot3=6\\z=5k=5\cdot3=15\end{matrix}\right.\)

`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}`

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 $a=4$

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

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

a.

$\sqrt{x^2-2x+1}=2$

$\Leftrightarrow \sqrt{(x-1)^2}=2$

$\Leftrightarrow |x-1|=2$

$\Rightarrow x-1=\pm 2$

$\Leftrightarrow x=3$ hoặc $x=-1$ (đều thỏa mãn)

b. ĐKXĐ: $x\geq 1$ hoặc $x\leq -1$

PT \(\Rightarrow \left\{\begin{matrix} x\geq 0\\ x^2-1=x^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 1=0\end{matrix}\right.\) (vô lý)

Vậy pt vô nghiệm

c. ĐKXĐ: $x\geq 5$

PT $\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=4$

$\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4$

$\Leftrightarrow 2\sqrt{x-5}=4$

$\Leftrightarrow \sqrt{x-5}=2$

$\Leftrightarrow x=2^2+5=9$ (thỏa mãn)

a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)

\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)

\(\Leftrightarrow9x+7=16\)

\(\Leftrightarrow9x=9\)

hay x=1

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

1)

a) \(-\dfrac{4}{3}.\dfrac{5}{12}+\dfrac{1}{3}.\dfrac{5}{12}=\dfrac{5}{12}.\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)=\dfrac{5}{12}.\left(-1\right)=-\dfrac{5}{12}\)

b) \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right).\dfrac{28}{3}=3+\dfrac{1}{5}-\dfrac{45}{14}.\dfrac{28}{3}\)

\(=3+\dfrac{1}{5}-30=-27+\dfrac{1}{5}=-\dfrac{134}{5}\)

2)

a) \(2x+19=25\)

\(2x=25-19=6\)

\(x=3\)

b) \(-\dfrac{2}{9}x-\dfrac{1}{7}=\dfrac{4}{21}\)

\(-\dfrac{2x}{9}=\dfrac{4}{21}+\dfrac{1}{7}=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}.\left(-\dfrac{9}{2}\right)=-\dfrac{3}{2}\)

a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5