(x+1)+(x+2)+(x+3)=4x

x+1+x+2+x+3=4x

(x+x+x)+(1+2+3)=4x

x*3+6=4x

6=1*x(bớt cả hai vế đi 3*x)

x=6/1(Tìm thừa số)

x=6

Mik sẽ k cho bạn đó mik viết nhầm

a, 1,5 +|2x - 2/3| = 3/2

            |2x - 2/3| = 3/2 - 1,5 

            |2x - 2/3| = 0

<=> 2x - 2/3 = 0 

<=> 2x = 0 + 2/3

<=> 2x = 2/3

<=> x = 2/3 : 2

<=> x = 1/3 

Vậy x = 1/3

b, 3/4 - |1/4 - x| = 5/8

            |1/4 - x| = 3/4 - 5/8

            |1/4 - x| = 1/8

<=> 1/4 - x = 1/8

       1/4 - x = /1/8

<=> x = 1/4 - 1/8

       x = 1/4 - ( -1/8)

<=> x = 1/8

       x = 3/8

Vậy x thuộc { 1/8 ; 3/8 }

c, 3 + | x+2 | = 2

          | x+2 | = 2 - 3 

          | x+2 | = -1 

<=> x + 2 = -1

       x + 2 = 1

<=> x = -1 - 2

       x = 1 -2 

<=> x = -3

       x = -1

Vậy x thuộc { -3 ; -1 }

a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)

\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)

mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)

\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

b) Tương tự câu a, ta có:

\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)

c. Tương tự, ta có:

\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)

a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...

b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)

Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...

c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...

Chân thành cảm ơn các bạn!

bài nào hả bạn

Cái đè bài đâu ?

???? Sao không có đề bài nhỉ

\(\frac{3}{13}.\frac{5}{9}+\frac{1}{6}:\frac{13}{3}+1\)

\(=\frac{3}{13}.\frac{5}{9}+\frac{1}{6}.\frac{3}{13}+1\)

\(=\frac{3}{13}.\left(\frac{5}{9}+\frac{1}{6}\right)+1\)

\(=\frac{3}{13}.\left(\frac{30+9}{54}\right)+1\)

\(=\frac{3}{13}.\frac{39}{54}+1\)

\(=\frac{1}{6}+1\)

\(=\frac{7}{6}\)

\(\frac{5}{6}-\frac{7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)

\(=\frac{5}{6}-\frac{7}{9}.\left(\frac{2}{13}-\frac{11}{13}\right)+\frac{-2}{9}\)

\(=\frac{5}{6}-\frac{7}{9}.\frac{-9}{13}-\frac{2}{9}\)

\(=\frac{5}{6}-\frac{-7}{13}-\frac{2}{9}\)

\(\frac{5}{6}-\frac{7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)

\(=\frac{5}{6}-\frac{7}{9}.\left(\frac{2}{13}-\frac{11}{13}\right)+\frac{-2}{9}\)

\(=\frac{5}{6}-\frac{7}{9}.\frac{-9}{13}-\frac{2}{9}\)

\(=\frac{5}{6}-\frac{-7}{13}-\frac{2}{9}\)

\(=\frac{5}{6}+\frac{7}{13}-\frac{2}{9}\)

\(=\frac{195+126-52}{234}\)

\(=\frac{269}{234}\)

\(\frac{3}{13}.\frac{5}{9}+\frac{1}{6}:\frac{13}{3}+1\)

\(=\frac{3}{13}.\frac{5}{9}+\frac{1}{6}.\frac{3}{13}+1\)

\(=\frac{3}{13}.\left(\frac{5}{9}+\frac{1}{6}\right)+1\)

\(=\frac{3}{13}.\left(\frac{30+9}{54}\right)+1\)

\(=\frac{3}{13}.\frac{39}{54}+1\)

\(=\frac{1}{6}+1=\frac{1}{6}+\frac{6}{6}\)

\(=\frac{7}{6}\)

\(\frac{-7}{9}.\frac{2}{13}-\frac{7}{9}.\frac{11}{13}+\frac{-2}{9}\)

\(=\frac{-7}{9}.\frac{2}{13}+\frac{-7}{9}.\frac{11}{13}+\frac{-2}{9}\) 

\(=\frac{-7}{9}.\left(\frac{2}{13}+\frac{11}{13}\right)+\frac{-2}{9}\)

\(=\frac{-7}{9}.1+\frac{-2}{9}\)

\(=\frac{-7}{9}+\frac{-2}{9}\)

\(=\frac{-9}{9}=-1\)

\(\frac{2}{13}.\frac{2}{7}.5\)

\(=\frac{2.2.5}{13.7}\)

\(=\frac{20}{91}\)

\(\frac{1}{5}.\frac{11}{12}.\frac{21}{6}\)

\(=\frac{11.21}{5.12.6}\)

\(=\frac{231}{360}=\frac{77}{120}\)

Bài 1: 

b) Ta có: \(D=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)

\(=\dfrac{-5}{10}\cdot\dfrac{-4}{10}\cdot\dfrac{-3}{10}\cdot...\cdot0\cdot...\cdot\dfrac{3}{10}\cdot\dfrac{4}{10}\cdot\dfrac{5}{10}\)

=0

0,2-0,375+5/11/-0,3+9/16-15/22

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uk 276 nha mik tính là vậy còn ko biết đúng ko nữa cho mik 1 k nha hihi / HT/

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