\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)

a: =>6/x=x/24

=>x^2=144

=>x=12 hoặc x=-12

b: =>x(1-7/12+3/8)=5/24

=>x*19/24=5/24

=>x=5/24:19/24=5/19

c: =>(x-1/3)^2=1+3/4+1/2=9/4

=>x-1/3=3/2 hoặc x-1/3=-3/2

=>x=11/6 hoặc x=-7/6

d: =>(x-3)^2=16

=>x-3=4 hoặc x-3=-4

=>x=-1 hoặc x=7

e: =>9/x=-1/3

=>x=-27

f: =>x-1/2=0 hoặc -x/2-3=0

=>x=1/2 hoặc x=-6

Ta có:\(\left(x-3\right)^2-\left(x+3\right)^2=\left(x-3-x-3\right)\left(x-3+x+3\right)=-12x=24\)

\(\Rightarrow x=24\div\left(-12\right)=-2\)

Vậy x=-2

tào lao

(x-3)²bắt buộc phải nhỏ hơn (x+3)²mà (x-3)²-(x+3)²=24

xin thưa huynh van duong đó là đề cương trường mk đó

Bij loi a ban

3^x+2-3^x=24

<=> 3^x(9-1)=24

<=> 3^x=3=3¹

<=> x=1

Trả louwf ;.....................

x = 1.......................

Hk tốt.........................

3^x+2-3^x=24

3^x.3²-3^x=24

3^x.(3²-1)=24

3^x.(9-1)=24

3^x.8=24

3^x=24:8

3^x=3

x=1

vậy x=1

3^x+2 - 3 = 24

3^x x 3² - 3 ^x = 24

3^x x ( 9 -1) =24

3^x x 8 = 24

3^x = 24 : 8

3^x = 3

suy ra x = 3

Vậy x =3

\(3^{x+2}-3^x=24\)

\(\Rightarrow3^x.\left(3^2-1\right)=24\)

\(\Rightarrow3^x.8=24\)

\(\Rightarrow3^x=3\)

\(\Rightarrow x=1\)

a) 4(2x+7)-3(3x-2)=24

8x+28-9x-6=24

(8x-9x)+(28-6)=24

(-1)x+22=24

(-1)x=24-22=2

x=2:(-1)=-2

b) -2(x+3)+(-4)²=3(1-x)

(-2)x+(-6)+16=3-3x

(-2)x+(-6)+16+3x=3

[(-2)x+3x]+[(-6)+16]=3

x+10=3

x=3-10=-7

\(\Rightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Rightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)=24\)

Đặt \(x^2+3x+1=t\)

\(\Rightarrow\left(t-1\right)\left(t+1\right)=24\)

\(\Rightarrow t^2-1=24\Rightarrow t^2=25\Rightarrow t=5;-5\)

Xét t=5 thì \(x^2+3x+1=5\Rightarrow x^2+3x-4=0\)

\(\Rightarrow x^2-x+4x-4=0\)

\(\Rightarrow x\left(x-1\right)+4\left(x-1\right)=0\)

\(\Rightarrow\left(x+4\right)\left(x-1\right)=0\Rightarrow x=-4;1\)

Xét t=-5 ta có

\(x^2+3x+1=-5\Rightarrow x^2+3x+6=0\)

\(\Rightarrow x_1=\frac{-3+\sqrt{15}i}{2};x_2=\frac{-3-\sqrt{15}i}{2}\)

mà \(x\in Z\)nên x=-4;1

x^2 -2x = 24

=> x^2 - 2x - 24=0

=>x^2 -8x+6x - 24 = 0

=> ( x^2- 8x)+( 6x-24) = 0

=> x(x-8) + 6(x-8) = 0

=> (x+6)(x-8)=0

=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)

\(=\frac{\left(2.5\right)^4.3^4-2^4\left(3.5\right)^2}{2^8.5^2.3^3}=\frac{2^4.3^2.5^2\left(5^2.3^2-1\right)}{2^8.5^2.3^3}=\frac{255-1}{16.3}=\frac{14}{3}\)

xin lỗi mình làm lộn bài

x=6

\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\)

\(\Rightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\)

\(\Rightarrow5x^2-13x+3=-10x^2-15x-24\)

\(\Rightarrow5x^2+10x^2-13x+15x+3+24=0\)

\(\Rightarrow15x^2+2x+27=0\)

Ta có: 

\(\Delta=2^2-4\cdot15\cdot27==-1616< 0\)

Nên pt vô nghiệm

\(5x\left(x-2\right)-3\left(x-1\right)=20x^2-15x\left(2x+1\right)-24\\ \Leftrightarrow5x^2-10x-3x+3=20x^2-30x^2-15x-24\\ \Leftrightarrow5x^2-20x^2+30x^2-10x-3x+15x+3+24=0\\ \Leftrightarrow15x^2+2x+27=0\\ \Leftrightarrow15x^2-2.x.\sqrt{15}+\dfrac{2}{15}+\dfrac{403}{15}=0\\ \Leftrightarrow\left(\sqrt{15}x+\dfrac{\sqrt{30}}{15}\right)^2+\dfrac{403}{15}=0\left(Vô.lí\right)\\ Vậy:Không.có.x.thoả\)