a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)

\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)

\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)

Đến đây tự làm tiếp nhé

b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

=> x = 75, y = 50, z = 30

c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)

\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)

\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)

=> x=... , y=... , z=...

d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)

Ta có: xy = 90 => 2k.5k = 90 => 10k² = 90 => k² = 9 => k = 3 hoặc -3

Với k = 3 => x = 6, y = 15

Với k = -3 => x = -6, y = -15

Vậy...

e, Tương tự câu d

b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)

=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)

     \(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)

      \(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)

a)\(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)

\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\)

\(\dfrac{2}{5}+x=\dfrac{3}{12}\)

\(\dfrac{2}{5}+x=\dfrac{1}{4}\)

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

\(x=\dfrac{5}{20}-\dfrac{8}{20}\)

\(x=\dfrac{-3}{20}\)

b)\(2x\left(x-\dfrac{1}{7}\right)=0\)

\(\Rightarrow2x=0\) hoặc \(x-\dfrac{1}{7}=0\)

\(x=0:2\) \(x=0+\dfrac{1}{7}\)

\(x=0\) \(x=\dfrac{1}{7}\)

\(\Rightarrow x=0\) hoặc \(x=\dfrac{1}{7}\)

c)\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

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

\(\dfrac{1}{4}:x=\dfrac{8}{20}-\dfrac{15}{20}\)

\(\dfrac{1}{4}:x=\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\)

\(x=\dfrac{1}{4}.\dfrac{-20}{7}\)

x= \(\dfrac{1.\left(-5\right)}{1.7}\)

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

a) Ta có :\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)

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

\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)\(=\dfrac{5x+2y+z}{555+444+333}=\dfrac{1100}{1332}=\dfrac{275}{333}\)

Từ đó tìm được x;y;z

b) Từ \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)

Đặt \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4k\\y^2=9k\end{matrix}\right.\)

\(\Rightarrow x^2\cdot y^2=4k\cdot9k=52\)

\(\Rightarrow36k^2=52\)

\(\Rightarrow k^2=\dfrac{13}{9}\) (sai đề)

b: Sửa đề: x^2+y^2=52

Đặt x/2=y/3=k

=>x=2k; y=3k

x^2+y^2=52

=>4k^2+9k^2=52

=>k^2=4

TH1: k=2

=>x=4; y=6

TH2: k=-2

=>x=-4; y=-6

c: Đặt x/5=y/3=k

=>x=5k; y=3k

x^2-y^2=16

=>25k^2-9k^2=16

=>k^2=1

TH1: k=1

=>x=5; y=3

TH2: k=-1

=>x=-5; y=-3

d: Đặt x/2=y/3=k

=>x=2k; y=3k

Ta có: xy=54

=>2k*3k=54

=>6k^2=54

=>k^2=9

TH1: k=3

=>x=6; y=9

TH2: k=-3

=>x=-6; y=-9

e: Đặt x/4=y/3=k

=>x=4k; y=3k

Ta có: xy=12

=>4k*3k=12

=>k^2=1

TH1: k=1

=>x=4; y=3

TH2: k=-1

=>x=-4; y=-3

a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)

Lập bảng xét dấu:

TH : Xét x < -2

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

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

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

-2x = 2\(\dfrac{1}{4}\)

=> x = \(-1\dfrac{1}{8}\) ( loại )

TH 2: \(-2\le x< \dfrac{1}{2}\)

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

=> \(2,5=\dfrac{3}{4}\) ( loại )

TH3 : \(x\ge\dfrac{1}{2}\)

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

2x + 1,5 = \(\dfrac{3}{4}\)

x = -0,375( loại )

vậy ....

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)

c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)

TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)

\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)

TH2 : \(x-1< 0\Rightarrow x< 1\)

=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)

Vậy x = 1

a) Ta có : \(x - 2xy + y - 3 = 0\)

\(\Rightarrow-2xy+x+y=3\)

\(\Rightarrow-2.\left(-2xy+x+y\right)=-2.3\)

\(\Rightarrow4xy-2x-2y=-6\)

\(\Rightarrow4xy-2x-2y+1=-6+1\)

\(\Rightarrow2x.\left(2y-1\right).\left(2y-1\right)=-5\)

\(\Rightarrow\left(2y-1\right).\left(2x-1\right)=-5=1.\left(-5\right)=-5.1=\left(-1\right).5=5.\left(-1\right)\)

Tự lập bảng đi -.-

Vậy có 5 bộ số (x, y, z) thoã mãn: (0,0,0); (3,2,6);(-3,-2,6);(3,-2,-6);(-3,2.-6)

Lời giải:

\(x^3y^2(xy^2)=x^3.x.y^2.y^2=x^4y^4\)

\(-3x^3y.\frac{1}{5}x^2y=\frac{-3}{5}x^3.x^2.y.y=\frac{-3}{5}x^5y^2\)

\(\frac{2}{5}x^3\frac{1}{2}(xy)^2=\frac{1}{5}x^3.x^2.y^2=\frac{1}{5}x^5y^2\)

\(\frac{1}{2}(xy)^2\frac{2}{5}(xy)^2=\frac{1}{5}x^2.x^2.y^2.y^2=\frac{1}{5}x^4y^4\)

Vậy các đơn thức phần a,b,c đồng dạng với nhau; đơn thức d và e đồng dạng với nhau.