Bài 1: 

Ta có: \(3x=2y\)

nên \(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=-15

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)

Vậy: (x,y)=(-6;-9)

Bài 2: 

a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x+y-z=20

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

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)

Vậy: (x,y,z)=(40;30;50)

\(a,x-\dfrac{2}{3}=\dfrac{1}{6}\\ \Rightarrow x=\dfrac{5}{6}\\ b,2x+\dfrac{1}{2}=-\dfrac{5}{3}\\ \Rightarrow2x=-\dfrac{13}{6}\\ \Rightarrow x=-\dfrac{13}{12}\\ c,3x+\dfrac{3}{2}=x-\dfrac{5}{3}\\ \Rightarrow-4x+\dfrac{3}{2}=-\dfrac{5}{3}\\ \Rightarrow-4x=-\dfrac{19}{6}\\ \Rightarrow4x=\dfrac{19}{6}\\ \Rightarrow x=\dfrac{19}{24}.\)

a) x - 2/3 = 1/6

x = 1/6 + 2/3

x = 5/6

b) 2x + 1/2 = -5/3

2x = -5/3 - 1/2

2x = -13/6

x = -13/6 : 2

x = -13/12

c) 3x + 3/2 = x - 5/3

3x - x = -5/3 - 3/2

2x = -19/6

x = -19/6 : 2

x = -19/12

\(\left|x+\dfrac{1}{3}\right|-4=-1\)

\(\Rightarrow\left|x+\dfrac{1}{3}\right|=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)

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

\(\Rightarrow\left|x+\dfrac{1}{3}\right|=-1+4=3\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)

Vậy..........

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

\(\Rightarrow1\dfrac{3}{4}x=-\dfrac{4}{5}-1\dfrac{1}{2}=\dfrac{-23}{10}\)

\(\Rightarrow x=\dfrac{-23}{10}:1\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{-46}{35}\)

a) \(-\dfrac{3}{5}-x=-0,75\)

\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)

b) \(1\dfrac{4}{5}=-0,15-x\)

\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)

\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)

\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)

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

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

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

a) \(-\dfrac{3}{5}-x=-0,75\)

\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)

\(x=\dfrac{27}{20}\)

________

b) \(1\dfrac{4}{5}=-0,15-x\)

\(=>-0,15-x=\dfrac{9}{5}\)

\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)

\(x=\dfrac{-39}{20}\)

c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)

\(x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)

\(x=\dfrac{6}{15}=\dfrac{2}{5}\)

Sưa đề:: \(\frac{x+4}{20}+\frac{x+3}{21}=\frac{x+2}{22}+\frac{x+1}{23}\)

=>\(\frac{x+4}{20}+1+\frac{x+3}{21}+1=\frac{x+2}{22}+1+\frac{x+1}{23}+1\)

=>\(\frac{x+24}{20}+\frac{x+24}{21}=\frac{x+24}{22}+\frac{x+24}{23}\)

=>\(\left(x+24\right)\left(\frac{1}{20}+\frac{1}{21}-\frac{1}{22}-\frac{1}{23}\right)=0\)

=>x+24=0

=>x=-24

a) \(4^{x+2}.3^x=16.12^5\)

\(\Leftrightarrow4^{x+2}.3^x=4^2.4^5.3^5\)

\(\Leftrightarrow4^{x-5}.3^{x-5}=1\)

\(\Leftrightarrow12^{x-5}=1\)

\(\Leftrightarrow x-5=0\Leftrightarrow x=5\)

\(a)4^{x+2}.3^x=4^2.\left(4.3\right)^5\)

\(4^{x+2}.3^x=4^7.3^5\)

\(\Rightarrow4^{x+2}=4^7;3^x=3^5\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

b) \(\dfrac{4}{14}-\dfrac{1}{5}+\dfrac{11}{15}-\dfrac{24}{5}\)

\(=-\dfrac{1}{5}-\dfrac{24}{5}+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-\left(\dfrac{1}{5}+\dfrac{24}{5}\right)+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-\left(\dfrac{1+24}{5}\right)+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-\dfrac{25}{5}+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=-5+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=\dfrac{-70}{14}+\dfrac{4}{14}+\dfrac{11}{15}\)

\(=\dfrac{-70+4}{14}+\dfrac{11}{15}\)

\(=\dfrac{-\left(70-4\right)}{14}+\dfrac{11}{15}\)

\(=\dfrac{-66}{14}+\dfrac{11}{15}\)

\(=\dfrac{-990}{210}+\dfrac{154}{210}\)

\(=\dfrac{-990+154}{210}\)

\(=\dfrac{-\left(990-154\right)}{210}\)

\(=\dfrac{-836}{210}\)

\(=\dfrac{-418}{105}\)

-Đang xy sao thành ab vậy bạn?

Bài 1:

a) \(\left|3x-5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{1}{3};x_2=3\)

b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

cho đáp án tự làm (vì cách lm của mik bị ném đá khá nhiều lần òi :D)

\(x=-1\)

c) như câu b nhé :D

\(x=-2004\)

thank bạn!!

Bài 1:

Vì \(\dfrac{a}{b}< \dfrac{c}{d}\) nên ad<bc (1)

Xét tích; a.(b+d)=ab+ad (2)

b.(a+c)=ba+bc (3)

Từ (1),(2),(3) suy ra a.(b+d)<b.(a+c) .

Do đó \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (4)

Tương tự ta lại có \(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) (5)

Kết hợp (4),(5) => \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)

hay x<y<z

​Bài 2:

a) x là một số hữu tỉ \(\Leftrightarrow\)\(b-15\ne0\Leftrightarrow b\ne15\)

b)x là số hữu tỉ dương\(\Leftrightarrow b-15>0\Leftrightarrow b>15\)

c) x là số hữu tỉ âm \(\Leftrightarrow b-15< 0\Leftrightarrow b< 15\)

Bài 3:

Ta có: \(\left|x-\dfrac{1}{3}\right|\ge0\) (dấu bằng xảy ra \(\Leftrightarrow x=\dfrac{1}{3}\))

=>\(\left|x-\dfrac{1}{3}\right|+\dfrac{1}{4}\ge\dfrac{1}{4}>\dfrac{1}{5}\)

Vậy A\(>\dfrac{1}{5}\)

​Bài 4:

M>0 \(\Leftrightarrow x+5;x+9\) cùng dấu.Ta thấy x+5<x+9 nên chỉ có 2 trường hợp

M>0 \(\left[{}\begin{matrix}x+5;x+9\left(duong\right)\\x+5;x+9\left(am\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+5\ge0\\x+9\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge-5\\x\ge-9\end{matrix}\right.\)

​Bài 5:

Ta dùng phương pháp phản chứng:

Giả sử tồn tại 2 số hữu tỉ x và y thỏa mãn đẳng thức \(\dfrac{1}{x+y}=\dfrac{1}{x}+\dfrac{1}{y}\)

=>\(\dfrac{1}{x+y}=\dfrac{x+y}{x.y}\Leftrightarrow\left(x+y\right)^2=x.y\)

Đẳng thức này không xảy ra vì \(\left(x+y\right)^2>0\) còn x.y <0 ( do x,y là 2 số trái dấu,không đối nhau)

Vậy không tồn tại 2 số hữu tỉ x và y trái dấu ,không đối nhau thỏa mãn đề bài