1. Áp dụng TCDTSBN ta có:

$\frac{x-1}{3}=\frac{y-2}{4}=\frac{z+5}{6}=\frac{x-1+(y-2)-(z+5)}{3+4-6}$

$=\frac{x+y-z-8}{1}=\frac{8-8}{1}=0$

$\Rightarrow x-1=y-2=z+5=0$

$\Rightarrow x=1; y=2; z=-5$

2.

Có:

$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}=\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}$

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

$\frac{x+1}{2}=\frac{y+3}{4}=\frac{z+5}{6}=\frac{2x+2}{4}=\frac{3y+9}{12}=\frac{4z+20}{24}=\frac{2x+2+3y+9+4z+20}{4+12+24}=\frac{2x+3y+4z+31}{40}=\frac{9+31}{40}=1$

Suy ra:

$x+1=2.1=2\Rightarrow x=1$

$y+3=1.4=4\Rightarrow y=1$

$z+5=6.1=6\Rightarrow z=1$

$

\(e,112-45+5x=87\)

\(67+5x=87\)

\(5x=20\)

\(x=4\)

\(f,6^2+64:\left(x-1\right)=52\)

\(36+64:\left(x+1\right)=52\)

\(64:\left(x+1\right)=16\)

\(x+1=4\)

\(x=3\)

\(g,\left(55-x\right)\left(2x-44\right)=0\)

\(\Rightarrow\hept{\begin{cases}55-x=0\\2x-44=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=55\\x=22\end{cases}}\)

Vậy ....

\(h,x^2+4=13\)

\(x^2=9\)

\(x^2=\pm3^2\)

\(\Rightarrow x=\pm3\)

a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)

x254n3jsm3,s3333

a: \(\frac{52}{17}>\frac{51}{17}=3\)

\(3=\frac{121}{41}>\frac{120}{41}\)

Do đó: \(\frac{52}{17}>\frac{120}{41}\)

b: \(\frac34+\frac14:\left(\frac{7}{12}-\frac16\right)\)

\(=\frac34+\frac14:\left(\frac{7}{12}-\frac{2}{12}\right)\)

\(=\frac34+\frac14:\frac{5}{12}\)

\(=\frac34+\frac14\times\frac{12}{5}=\frac34+\frac35=\frac{15}{20}+\frac{12}{20}=\frac{27}{20}\)

c: \(372,463\cdot998+744,926\)

\(=372,463\cdot998+372,463\cdot2\)

\(=372,463\times\left(998+2\right)=372,463\times1000=372463\)

d: Số số hạng trong dãy số 2;4;6;...;100 là:

\(\left(100-2\right):2+1=98:2+1=49+1=50\) (số)

\(2-4+6-8+10-12+\cdots+98-100+102\)

\(=\left(2-4\right)+\left(6-8\right)+\cdots+\left(98-100\right)+102\)

=(-2)+(-2)+...+(-2)+102

\(=-2\cdot\frac{50}{2}+102=-50+102=52\)

e: (y+112)-113=79

=>y+112-113=79

=>y-1=79

=>y=79+1=80

f: \(\frac34-y=\frac12\)

=>\(y=\frac34-\frac12=\frac14\)

g: \(\left(\frac45-2\times y\right)+\frac16=\frac56\)

=>\(\frac45-2\times y=\frac56-\frac16=\frac46=\frac23\)

=>\(2\times y=\frac45-\frac23=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}\)

=>\(y=\frac{2}{15}:2=\frac{1}{15}\)

h: (y+1)+(y+2)+...+(y+50)=1750

=>50y+(1+2+...+50)=1750

=>\(50y+50\times\frac{51}{2}=1750\)

=>50y+1275=1750

=>50y=1750-1275=475

=>\(y=\frac{475}{50}=9,5\)

E, F, G, H, I tí nữa Thầy rảnh Thầy giải giúp nhé!

\(x:y=1\dfrac{2}{3}\Rightarrow\dfrac{x}{y}=\dfrac{5}{3}\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}\)

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

\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{60}{2}=30\)

\(\dfrac{x}{5}=30\Rightarrow x=150\\ \dfrac{y}{3}=30\Rightarrow y=90\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2+y^2}{4+9}=\dfrac{52}{13}=4\)

​\(\dfrac{x^2}{4}=4\Rightarrow x^2=16\\ \Rightarrow\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)​

\(\dfrac{y^2}{9}=4\Rightarrow y^2=36\Rightarrow\left[{}\begin{matrix}y=-6\\y=6\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left\{\left(-4;-6\right);\left(4;6\right)\right\}\)

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)

Áp dụng t.c của dãy tỉ só bằng nhau,ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{x+y}{3+4}=\dfrac{16}{7}\)

=>\(x=\dfrac{16}{7}.3=\dfrac{48}{7}\)

\(y=\dfrac{16}{7}.4=\dfrac{64}{7}\)

\(z=\dfrac{16}{7}.5=\dfrac{80}{7}\)

Vậy...

Các câu sau tương tự

ADTC dãy tỉ số bằng nhau ta cs

\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=\frac{x^2+y^2}{2^2+3^2}=\frac{52}{13}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=4\\\frac{y}{3}=4\\\frac{z}{4}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\y=12\\z=16\end{matrix}\right.\)