1. a) Ta có: M  = |x + 15/19| \(\ge\)0 \(\forall\)x

Dấu "=" xảy ra <=> x + 15/19 = 0 <=> x = -15/19

Vậy MinM = 0 <=> x = -15/19

b) Ta có: N = |x  - 4/7| - 1/2 \(\ge\)-1/2 \(\forall\)x

Dấu "=" xảy ra <=> x - 4/7 = 0 <=> x = 4/7

Vậy MinN = -1/2 <=> x = 4/7

2a) Ta có: P = -|5/3 - x|  \(\le\)0 \(\forall\)x

Dấu "=" xảy ra <=> 5/3 - x = 0 <=> x = 5/3

Vậy MaxP = 0 <=> x = 5/3

b) Ta có: Q = 9 - |x - 1/10| \(\le\)9 \(\forall\)x

Dấu "=" xảy ra <=> x - 1/10 = 0 <=> x = 1/10

Vậy MaxQ = 9 <=> x = 1/10

\(a,\frac{-1}{2}+\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}.\)

\(\Rightarrow\left(x-3\right):\frac{-1}{2}=-1\frac{2}{3}-\frac{-1}{2}=\frac{-7}{6}\)

\(\Rightarrow x-3=\frac{-7}{6}\cdot\frac{-1}{2}=\frac{7}{12}\)

\(\Rightarrow x=\frac{7}{12}+3=3\frac{7}{12}\)

\(b.2,25+\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}\)

\(\Rightarrow\frac{3}{2}:\left(x-5\right)=2\frac{1}{2}-2,25=\frac{1}{4}\)

\(\Rightarrow x-5=\frac{3}{2}:\frac{1}{4}=6\)

\(\Rightarrow x=6+5=11\)

\(c,\left(\frac{1}{3}-x\right)^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2=\left(-\frac{1}{2}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=\frac{1}{2}\\\frac{1}{3}-x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}\\x=\frac{1}{3}-\frac{-1}{2}=\frac{5}{6}\end{cases}}\)

\(d,\frac{3}{2}+\frac{x-1}{3}=1\)

\(\Rightarrow\frac{x-1}{3}=1-\frac{3}{2}=-\frac{1}{2}\)

\(\Rightarrow x-1=-\frac{1}{2}\cdot3=-\frac{3}{2}\)

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

\(e,-\frac{6}{8}+\frac{x}{12}=\frac{5}{6}\)

\(\Rightarrow\frac{x}{12}=\frac{5}{6}-\frac{-6}{8}=\frac{19}{12}\)

\(\Rightarrow x=19\)

\(g,\frac{1}{2}-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}\)

\(\Rightarrow-\frac{1}{3}\left(x-2\right)=-\frac{2}{3}-\frac{1}{2}=-\frac{7}{6}\)

\(\Rightarrow x-2=\frac{-7}{6}:\frac{-1}{3}=\frac{7}{2}\)

\(\Rightarrow x=\frac{7}{2}+2=2\frac{7}{2}\)

\(h,\frac{5}{2}\left(x+1\right)-\frac{1}{2}=3\frac{1}{2}\)

\(\Rightarrow\frac{5}{2}\left(x+1\right)=3\frac{1}{2}-\frac{1}{2}=3\)

\(\Rightarrow x+1=3:\frac{5}{2}=\frac{6}{5}\)

\(\Rightarrow x=\frac{6}{5}-1=\frac{1}{5}\)

\(k,\frac{x}{3}-\frac{1}{2}=-2\left(x+1\right)+3\)

\(\Rightarrow x\cdot\frac{1}{3}-\frac{1}{2}=-2x-2+3\)

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

\(\Rightarrow\frac{7}{3}x=\frac{3}{2}\Rightarrow x=\frac{3}{2}:\frac{7}{2}=\frac{3}{7}\)

1.

a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)

b) x=0

d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)

e) \(x=\frac{2}{3}\)

\(a,\frac{15^3.\left(-5\right)^4}{\left(-3\right)^5.5^6}\)\(=\frac{3^3.5^3}{\left(-3\right)^5.5^2}\)\(=-\frac{5}{\left(3\right)^2}=-\frac{5}{9}\)

\(b,\frac{6^3.2.\left(-3\right)^2}{\left(-2\right)^9.3^7}\)\(=-\frac{6^3}{2^8.3^5}\)\(=-\frac{2^3.3^3}{2^8.3^5}\)\(=-\frac{1}{2^5.3^2}=-\frac{1}{288}\)

\(c,\frac{3^6.7^2-3^7.7}{3^7.21}\)\(=\frac{3^6.7\left(7-3\right)}{3^7.21}\)\(=\frac{3^6.7.4}{3^7.7.3}\)\(=\frac{4}{3.3}=\frac{4}{9}\)

\(a,\left(x-1,2\right)^2=4\)

\(\Rightarrow x-1,2=2\)

\(\Rightarrow x=3,2\)

\(b,\left(x+1\right)^3=-125\)

\(\Rightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Rightarrow x+1=-5\Rightarrow x=-6\)

\(c,\left(x-5\right)^3=2^6\)

\(\Rightarrow\left(x-5\right)^3=4^3\)

\(\Rightarrow x-5=4\Rightarrow x=9\)

\(d,\left(2x+1\right)^{x+1}=5^{x+1}\)

\(\Rightarrow2x+1=5\Rightarrow x=2\)

Bài 2:

1)

a) \(\frac{3}{5}-x=25\%\)

=> \(\frac{3}{5}-x=\frac{1}{4}\)

=> \(x=\frac{3}{5}-\frac{1}{4}\)

=> \(x=\frac{7}{20}\)

Vậy \(x=\frac{7}{20}.\)

b) \(0,16:x=x:36\)

=> \(\frac{0,16}{x}=\frac{x}{36}\)

=> \(0,16.36=x.x\)

=> \(x.x=\frac{144}{25}\)

=> \(x^2=\frac{144}{25}\)

=> \(\left[{}\begin{matrix}x=\frac{12}{5}\\x=-\frac{12}{5}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{12}{5};-\frac{12}{5}\right\}.\)

2)

a) Ta có: \(5x=7y.\)

=> \(\frac{x}{y}=\frac{7}{5}\)

=> \(\frac{x}{7}=\frac{y}{5}\) và \(y-x=18.\)

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

\(\frac{x}{7}=\frac{y}{5}=\frac{y-x}{5-7}=\frac{18}{-2}=-9.\)

\(\left\{{}\begin{matrix}\frac{x}{7}=-9=>x=\left(-9\right).7=-63\\\frac{y}{5}=-9=>y=\left(-9\right).5=-45\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(-63;-45\right).\)

b) Ta có: \(\frac{x}{y}=0,8.\)

=> \(\frac{x}{y}=\frac{4}{5}\)

=> \(\frac{x}{4}=\frac{y}{5}\) và \(x+y=18.\)

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

\(\frac{x}{4}=\frac{y}{5}=\frac{x+y}{4+5}=\frac{18}{9}=2.\)

\(\left\{{}\begin{matrix}\frac{x}{4}=2=>x=2.4=8\\\frac{y}{5}=2=>y=2.5=10\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(8;10\right).\)

Mình chỉ làm thế này thôi nhé.

Chúc bạn học tốt!

mik chỉ làm b3,2 thôi nha^_^

nhưng x là số gì

x ϵ z

\(\frac{4}{5}x+0=4,5\)

\(\frac{4}{5}x=4,5\)

\(x=4,5:\frac{4}{5}\)

\(x=5,625\)

vậy \(x=5,625\)

\(\frac{x}{3}=\frac{-5}{9}\)

\(\Rightarrow9x=-5.3\)

\(\Rightarrow9x=-15\)

\(\Rightarrow x=\frac{-5}{3}\)

vậy \(x=\frac{-5}{3}\)

\(\left|x+5\right|-\frac{1}{3}=\frac{2}{3}\)

\(\left|x+5\right|=\frac{2}{3}+\frac{1}{3}\)

\(\left|x+5\right|=1\)

\(\Rightarrow\orbr{\begin{cases}x+5=1\\x+5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-4\\x=-6\end{cases}}\)

                vậy \(\orbr{\begin{cases}x=-4\\x=-6\end{cases}}\)

\(\left(x-2\right)^3=-125\)

\(\left(x-2\right)^3=\left(-5\right)^3\)

\(\Rightarrow x-2=-5\)

\(\Rightarrow x=-3\)

vậy \(x=-3\)

Câu d:

-1\(\frac23\) - (|2\(x\)| + \(\frac56\)) = - 2

-\(\frac53\) - |2\(x\)| - \(\frac56\) = - 2

|2\(x\)| = - \(\frac53\) - \(\frac56\) + 2

|2\(x\)| = - \(\frac52\) + 2

|2\(x\)| = - \(\frac12\) (vô lí vì trị tuyệt đối của một số luôn là một số không âm)

Không có giá trị nào của x thỏa mãn đề bài.

x ∈ ∅

Câu a:

|\(x\) - 3| = \(x\) + 4

Vì |\(x\) - 3| ≥ 0 ∀ \(x\) nên \(x\) + 4 ≥ 0 ⇒ \(x\) ≥ - 4

Với -4 ≤ \(x\) ≤ 3 ta có:

-\(x\) + 3 = \(x\) + 4

\(x\) + \(x\) = -4 + 3

2\(x\) = -1

\(x=\frac{-1}{2}\)

Với x > 3 ta có:

x - 3 = x + 4

x - x = 3 + 4

0 = 7 (vô lí)

Vậy x = -1/2 là nghiện duy nhất của phương trình.

Vậy \(x\) = -1/2

a) \(\left|2x+\frac{3}{4}\right|=\frac{1}{2}\)

     \(\orbr{\begin{cases}2x+\frac{3}{4}=\frac{1}{2}\\2x+\frac{3}{4}=\frac{-1}{2}\end{cases}}\) =>   \(\orbr{\begin{cases}2x=\frac{1}{2}-\frac{3}{4}\\2x=\frac{-1}{2}-\frac{3}{4}\end{cases}}\)  =>   \(\orbr{\begin{cases}2x=\frac{-1}{4}\\2x=\frac{-5}{4}\end{cases}}\) =>   \(\orbr{\begin{cases}x=\frac{-1}{8}\\x=\frac{-5}{8}\end{cases}}\)

Vậy \(x=\left\{\frac{-1}{8},\frac{-5}{8}\right\}\)

b) \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{2\frac{1}{4}}\)= \(\frac{3x}{2,7}=\frac{\frac{1}{4}}{\frac{9}{4}}\)

=> \(3x.\frac{9}{4}=2,7.\frac{1}{4}\)=>  \(\frac{27x}{4}=\frac{27}{40}\)

\(27x.40=27.4\)

\(1080.x=108\)

             \(x=\frac{1}{10}\)

Vậy \(x=\frac{1}{10}\)

c) \(\left|x-1\right|+4=6\)

\(\left|x-1\right|=6-4\)

\(\left|x-1\right|=2\)

\(\orbr{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)=>  \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy \(x=\left[3,-1\right]\)

d) \(\frac{x}{3}=\frac{y}{5}=>\frac{y}{5}=\frac{x}{3}=>\frac{y-x}{5-3}=\frac{24}{2}=12\)

e) \(\left(x^2-3\right)^2=16\)

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

\(x^2=7=>x=\sqrt{7}\)

Vậy \(x=\sqrt{7}\)

f) \(\frac{3}{4}+\frac{2}{5}x=\frac{29}{60}\)

               \(\frac{2}{5}x=\frac{29}{60}-\frac{3}{4}\) 

               \(\frac{2}{5}x=-\frac{4}{15}\)

          \(x=-\frac{4}{15}:\frac{2}{5}=-\frac{4}{15}.\frac{5}{2}=-\frac{2}{3}\)

Vậy \(x=-\frac{2}{3}\)

g) \(\left(-\frac{1}{3}\right)^3.x=\frac{1}{81}\)

\(\left(-\frac{1}{27}\right).x=\frac{1}{81}\)

\(x=\left(-\frac{1}{27}\right):\frac{1}{81}=\left(-\frac{1}{27}\right).81=-3\)

Vậy \(x=-3\)

k)\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60}\)

\(\frac{2}{5}x=\frac{3}{4}-\frac{29}{60}\)

\(\frac{2}{5}x=\frac{4}{15}\)

      \(x=\frac{2}{5}-\frac{4}{15}=>x=\frac{2}{15}\)

Vậy \(x=\frac{2}{15}\)

I) \(\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)

\(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}\)

\(\frac{3}{5}x=\frac{5}{14}\)

\(x=\frac{5}{14}:\frac{3}{5}=\frac{5}{14}.\frac{5}{3}=\frac{25}{42}\)

Vậy \(x=\frac{25}{42}\)