1/3

-13/12<-19/18

14) (x - 2) . (x + 4) = 0

\(\Rightarrow\) x - 2 = 0 hoặc x + 4 = 0

Nếu x - 2 = 0

x = 0 + 2

x = 2

Nếu x + 4 = 0

x = 0 - 4

x = -4

Vậy x \(\in\) {2 ; -4)

15) (x - 2) . (x + 15) = 0

\(\Rightarrow\) x - 2 = 0 hoặc x + 15 = 0

Nếu x - 2 = 0

x = 0 + 2

x = 2

Nếu x + 15 = 0

x = 0 - 15

x = -15

Vậy x \(\in\) {-15 ; 2}

các thánh giúp em với! khổ tâm > bài khó quá mà chẳng ai trl !\

Akai Haruma

Nguyễn Huy Tú

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Nguyễn Thanh Hằng

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Nguyễn Thị Hồng Nhung

cô Bùi Thị Vân giúp em với !

xin cô giáo tick cho ai trả lời đúng câu hỏi của em ạ !

Trời ! bài toán lớp 6 ư !

tui ko biet lam 

toán lớp 6 à bạn mình năm nay mới lên lớp 5 thôi

a)
\(\left|x\right|-2\left|x\right|+3\left|x\right|=16+6\left|x\right|-19\)
\(\left|x\right|-2\left|x\right|+3\left|x\right|-6\left|x\right|=16-19\)
\(\left|x\right|.\left(1-2+3-6\right)=-3\)
\(\left|x\right|.\left(-4\right)=-3\)
\(\left|x\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)


b,
2.(|x| - 5) - 15 = 9
\(2.\left(\left|x\right|-5\right)=9+15\)
\(2.\left(\left|x\right|-5\right)=24\)
\(\left|x\right|-5=24:2\)
\(\left|x\right|-5=12\)
\(\left|x\right|=12+5\)
\(\left|x\right|=17\)
\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)

c,
|8 - 2x| + |4y - 16| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|8-2x\right|=0\\\left|4y-16\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}8-2x=0\\4y-16=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=8\\4y=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)


d,

|x - 14| + |2y - x| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|x-14\right|=0\\\left|2y-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-14=0\\2y-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)

2.Tìm x, y, z biết

a,
2.|3x| + |y + 3| + |z - y| = 0
\(\Rightarrow\left\{{}\begin{matrix}2.\left|3x\right|=0\\\left|y+3\right|=0\\\left|z-y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x\right|=0\\y+3=0\\z-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=0\\y=-3\\z=y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)

b, (x - 3y)2 + | y + 4|= 0
\(\Rightarrow\left\{{}\begin{matrix}\left(x-3y\right)2=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-4\right)\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

bài 1 a)

\(\frac{15}{7}\times\frac43+\frac63\)

=\(\frac{60}{21}+\frac63\)

=\(\frac{60}{21}+\frac{42}{21}\)

=\(\frac{102}{21}\)

=\(\frac{34}{7}\)

b)\(\frac{-13}{41}\times\frac{17}{11}\times\frac{11}{17}\)

=\(\frac{-13}{41}\times\left(\frac{17}{11}\times\frac{11}{17}\right)\)

=\(\frac{-13}{41}\times1\)

=\(\frac{-13}{41}\)

a. /3x-4/=16+[1/19+18/19]

   /3x-4/=16+1

   /3x-4/=17

    Suy ra 3x-4=17 hoặc -17

-Nếu 3x-4=17                                                    -Nếu 3x-4=-17

         3x=17+4                                                            3x   =-17+4

         3x=21                                                                 3x   =-13

           x=21:3                                                                x    =-13:3

           x=7                                                                     x     =-13/3

b./2x+1/=x+2

  Suy ra 2x+1=x+2 hoặc -[x+2]

-Nếu 2x+1=x+2                                                -Nếu 2x+1=-[x+2]

        2x     =x+2-1                                                      2x+1=-x+[-2]

        2x     =x+1                                                         2x       =-x+[-2]-1

        2x-x  =1                                                             2x-[- x]=-3

          x     =1                                                             3x       =-3            

                                                                                  x         =-3:3

                                                                                  x         =-1

\(a,12,3\cdot4,7+5,3\cdot12,3-1,4\cdot7,5+7,5\cdot5,4\)

\(=12,3\cdot\left(4,7+5,3\right)-7,5\cdot\left(1,4+5,4\right)\)

\(=12,3\cdot10-7,5\cdot6,8\)

\(=123-51\)

\(=72.\)

\(b,3\cdot x-1=\frac{3}{4}:\frac{15}{16}\)

\(\Leftrightarrow3\cdot x-1=\frac{3}{4}\cdot\frac{16}{15}\)

\(\Leftrightarrow3\cdot x-1=\frac{4}{5}\)

\(\Leftrightarrow3\cdot x=\frac{4}{5}+1\)

\(\Leftrightarrow3\cdot x=\frac{9}{5}\)

\(\Leftrightarrow x=\frac{9}{5}\div3\)

\(\Leftrightarrow x=\frac{9}{5}\cdot\frac{1}{3}\)

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

a) \(12,3\times4,7+5,3\times12,3-1,4\times7,5+7,5\times5,4\)

\(=12,3\times\left(4,7+5,3\right)+7,5\times\left(5,4-1,4\right)\)

\(=12,3\times10+7,5\times4\)

\(=123+30\)

\(=153\)

b) \(3.x-1=\frac{3}{4}:\frac{15}{16}\)

\(\Rightarrow3.x-1=\frac{4}{5}\)

\(\Rightarrow3.x=\frac{9}{5}\Rightarrow x=\frac{3}{5}\)

\(4^{15}\cdot9^{15}\le2^x\cdot3^x\le18^{16}\cdot2^{16}\)

\(\Rightarrow\left(2^2\right)^{15}\cdot\left(3^2\right)^{15}\le2^x\cdot3^x\le\left(2\cdot3^2\right)^{16}\cdot2^{16}\)

\(\Rightarrow2^{30}\cdot3^{30}\le2^x\cdot3^x\le2^{32}\cdot3^{32}\)

\(\Rightarrow x=31\)