Câu 2:

a: 10km=10000m

10000m dây đồng có cân nặng là:

\(47:5\cdot10000=94000\left(g\right)\)

b: 300g=0,3kg=0,003 tạ

0,003 tạ nặng:

\(2,5:1\cdot0,003=\dfrac{3}{400}\left(kg\right)\)

Câu 1:

a:

\(\left|1-2x\right|>=0\forall x\)

=>\(3\left|1-2x\right|>=0\forall x\)

=>\(3\left|1-2x\right|-5>=-5\forall x\)

=>\(A>=-5\forall x\)

Dấu '=' xảy ra khi 1-2x=0

=>2x=1

=>x=1/2

Vậy: \(A_{Min}=-5\) khi x=1/2

b: \(2x^2>=0\forall x\)

=>\(2x^2+1>=1\forall x\)

=>\(\left(2x^2+1\right)^4>=1^4=1\forall x\)

=>\(\left(2x^2+1\right)^4-3>=1-3=-2\forall x\)

=>B>=-2\(\forall\)x

Dấu '=' xảy ra khi x=0

c: \(\left|x-\dfrac{1}{2}\right|>=0\forall x\)

\(\left(y+2\right)^2>=0\forall y\)

Do đó: \(\left|x-\dfrac{1}{2}\right|+\left(y+2\right)^2>=0\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\y+2=0\end{matrix}\right.\)

=>x=1/2 và y=-2

Bài 2:

1: \(\dfrac{x}{12}-\dfrac{5}{6}=\dfrac{1}{12}\)

=>\(\dfrac{x}{12}=\dfrac{1}{12}+\dfrac{5}{6}=\dfrac{1}{12}+\dfrac{10}{12}=\dfrac{11}{12}\)

=>x=11

2: \(\dfrac{2}{3}-1\dfrac{4}{15}x=-\dfrac{3}{5}\)

=>\(\dfrac{2}{3}-\dfrac{19}{15}x=-\dfrac{3}{5}\)

=>\(\dfrac{19}{15}x=\dfrac{2}{3}+\dfrac{3}{5}=\dfrac{10+9}{15}=\dfrac{19}{15}\)

=>\(x=\dfrac{19}{15}:\dfrac{19}{15}=1\)

3: \(\dfrac{\left(-3\right)^x}{81}=-27\)

=>\(\left(-3\right)^x=\left(-3\right)^3\cdot\left(-3\right)^4=\left(-3\right)^7\)

=>x=7

4: \(\left|x+0,237\right|=0\)

=>x+0,237=0

=>x=-0,237

5: \(\left(x-1\right)^2=25\)

=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

6: \(\left|2x-1\right|=5\)

=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

7: \(\left(x-1\right)^3=-\dfrac{8}{27}\)

=>\(\left(x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)

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

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

8: \(1\dfrac{2}{3}:\dfrac{x}{4}=6:0,3\)

=>\(\dfrac{5}{3}:\dfrac{x}{4}=20\)

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

=>3x=20/20=1

=>\(x=\dfrac{1}{3}\)

9: \(2\dfrac{2}{3}:x=1\dfrac{7}{9}:2\dfrac{2}{3}\)

=>\(\dfrac{\dfrac{8}{3}}{x}=\dfrac{\dfrac{16}{9}}{\dfrac{8}{3}}\)

=>\(\dfrac{16}{9}\cdot x=\dfrac{8}{3}\cdot\dfrac{8}{3}=\dfrac{64}{9}\)

=>16x=64

=>x=64/16=4

Bài 3:

1: Ta có: x-24=y

=>x-y=24

mà \(\dfrac{x}{7}=\dfrac{y}{3}\)

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

\(\dfrac{x}{7}=\dfrac{y}{3}=\dfrac{x-y}{7-3}=\dfrac{24}{4}=6\)

=>\(x=6\cdot7=42;y=6\cdot3=18\)

2: \(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}\)

mà x-y=48

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

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{x-y}{5-7}=\dfrac{48}{-2}=-24\)

=>\(x=-24\cdot5=-120;y=-24\cdot7=-168;z=-24\cdot2=-48\)

3: \(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}\)

mà x-y=4009 

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

\(\dfrac{x-1}{2005}=\dfrac{3-y}{2006}=\dfrac{x-1+3-y}{2005+2006}=\dfrac{4009+2}{4011}=1\)

=>\(x-1=2005;3-y=2006\)

=>x=2005+1=2006; y=3-2006=-2003

5: \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}\)

mà 2x+3y-z=-14

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

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{2\cdot3+3\cdot5-7}=\dfrac{-14}{14}=-1\)

=>\(x=-3;y=-5;z=-7\)

Bạn tách ra từng CH khác nhau đi nhé. Gộp 1 trong tất cả rất khó nhìn và lâu.

Help me pls

c.\(\dfrac{3}{7}+\dfrac{5}{7}:x=\dfrac{1}{3}\)

\(\dfrac{5}{7}:x=\dfrac{1}{3}-\dfrac{3}{7}\)

\(\dfrac{5}{7}:x=-\dfrac{2}{21}\)

\(x=\dfrac{5}{7}:-\dfrac{2}{21}\)

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

d.\(3\dfrac{1}{4}:\left|2x-\dfrac{5}{12}\right|=\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=3\dfrac{1}{4}:\dfrac{39}{16}\)

\(\left|2x-\dfrac{5}{12}\right|=\dfrac{4}{3}\)

\(\rightarrow\left[{}\begin{matrix}2x-\dfrac{5}{12}=\dfrac{4}{3}\\2x-\dfrac{4}{12}=-\dfrac{4}{3}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}2x=\dfrac{7}{4}\\2x=-\dfrac{11}{12}\end{matrix}\right.\) \(\rightarrow\left[{}\begin{matrix}x=\dfrac{7}{8}\\x=-\dfrac{11}{24}\end{matrix}\right.\)

A, \(\dfrac{4}{9}+x=\dfrac{5}{3}\)

\(x\)\(=\dfrac{5}{3}-\dfrac{4}{9}\)

\(x\)\(=\dfrac{11}{9}\)

B,\(\dfrac{3}{4}.x=\dfrac{-1}{2}\)

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

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

a)

\(\frac{4}{9} + x = \frac{5}{3}\)

=> \(x = \frac{5}{3}-\frac{4}{9}\)

=> \(x = \) \(\frac{11}{9}\)

Vậy \(x = \dfrac{11}{9}\)

b) 

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

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

=> \(x = \dfrac{-2}{3}\)

Vậy \(x = \dfrac{-2}{3}\)

c)

\( \dfrac{3}{7}+ \dfrac{5}{7}:x = \dfrac{1}{3}\)

=> \(\dfrac{5}{7}:x = \dfrac{1}{3}-\) \( \dfrac{3}{7}\)

=> \(\dfrac{5}{7}:x = \dfrac{-2}{21}\)

=> \(x = \dfrac{5}{7}:\dfrac{-2}{21}\)

=> \(x = \dfrac{-15}{2}\)

Vậy \(x = \dfrac{-15}{2}\)

d) 

\(3\dfrac{1}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \(\dfrac{13}{4} : |2x - \dfrac{5}{12} | = \dfrac{39}{16}\)

=> \( |2x - \dfrac{5}{12} | =\dfrac{13}{4} : \dfrac{39}{16}\)

=> \(|2x-\dfrac{5}{12} |= \dfrac{4}{3}\)

=> \(\left[\begin{matrix} 2x - \dfrac{5}{12} = \dfrac{4}{3}\\ 2x - \dfrac{5}{12} = \dfrac{4}{3}\end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{-4}{3}+\dfrac{5}{12}\\ 2x = \dfrac{-4}{3}+\dfrac{5}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} 2x = \dfrac{7}{4}\\ 2x = \dfrac{-11}{12} \end{matrix}\right.\)

=> \(\left[\begin{matrix} x = \dfrac{7}{8}\\ x = \dfrac{-11}{24} \end{matrix}\right.\)

Vậy \(x \in \) { \(\dfrac{7}{8} ; \dfrac{-11}{24}\) }

a × 9 - b × 9

= 9 × (a - b)

= 9 × 8/3

= 24

cảm ơn bạn

Chọn A

Áp dụng công thức tỉ lệ phân số ta có : 

\(\dfrac{a}{b}=\dfrac{c}{d}\)

\(\Rightarrow\dfrac{a^2}{b^2}=\dfrac{c^2}{d^2}=\dfrac{ac}{bd}\)

\(\dfrac{1}{a}-\dfrac{1}{b}=1\)

\(\Leftrightarrow\dfrac{b-a}{ab}=1\)

\(\Leftrightarrow b-a=ab\)

\(\Leftrightarrow a+ab-b=0\)

Lời giải:
a.

$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$

b.

$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$

$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$

c.

$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$

d.

$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$

$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$

e.

$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$

$\frac{-19}{15}: x=1$

$x=\frac{-19}{15}:1 =\frac{-19}{15}$

f.

$(-\frac{3}{4}+x).2\frac{2}{3}=1$

$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$

$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$