\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.\)

\(=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1\)

\(=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1\)

\(=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1\)

\(=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1\)

\(=x\left(\frac{5+3-8}{30}\right)+1\)

\(=x\cdot0+1=1\)

\(\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1\)

trả lời:

  \frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0.61​x+101​x−154​x+1+0.

=\left(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x\right)+1=(61​x+101​x−154​x)+1

=x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=x(61​+101​−154​)+1

=x\left(\frac{1\cdot5}{30}+\frac{1\cdot3}{30}-\frac{4\cdot2}{30}\right)+1=x(301⋅5​+301⋅3​−304⋅2​)+1

=x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=x(305​+303​−308​)+1

=x\left(\frac{5+3-8}{30}\right)+1=x(305+3−8​)+1

=x\cdot0+1=1=x⋅0+1=1

\Rightarrow\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1+0=1⇒61​x+101​x−154​x+1+0=1

ko chắc chúc bạn học tốt.

Bài làm

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=40\)

\(x.\frac{1}{30}+1=0\)

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

\(x=-1:\frac{1}{3}\)

\(x=-1.3\)

\(x=-3\)

# Học tốt #

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(\Rightarrow x.0+1=0\)

=> 1=0 ( Vô lý )

Vậy \(x\in\varnothing\)

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow x.\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)=1\)

\(\Rightarrow x.0=1\Rightarrow x=0\)

Vậy x=0

\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)

TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)

TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)

\(\Rightarrow x\in\left\{2;3;-4\right\}\)

\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)

\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)

\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)

a) (1/7x - 2/7)(-1/5x + 3/5)(1/3x + 4/3) = 0

3 trường hợp:

TH1: 1/7x - 2/7 = 0 <=> 1/7x = 0 + 2/7 <=> 1/7x = 2/7 <=> x = 2.7/7 = 2

=> x = 2

TH2: -1/5x + 3/5 = 0 <=> -1/5x = 0 - 3/5 <=> -1/5x = -3/5 <=> x = (-3/5).(-5) = 3

=> x = 3

TH3: 1/3x + 4/3 = 0 <=> 1/3x = 0 - 4/3 <=> 1/3x = -4/3 <=> x = x = 3.(-4/3) = -4

=> x = -4

Vậy: x = 2, 3, -4

b) 1/6x + 1/10x - 4/15x + 1 = 0

<=> 1/6x + 1/10x - 4/15x = 0 - 1

<=> 1/6x + 1/10x - 4/15x = -1

<=> 1/6x.30 + 1/10x.30 - 4/15x.30 = -1.30

<=> 5x + 3x - 8x = -30

<=> 0 = -30

=> không có x thỏa mãn

a, \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)

\(\Leftrightarrow-\frac{1}{10}x=-\frac{11}{10}\)

\(\Leftrightarrow x=11\)

b,\(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

\(\Leftrightarrow\frac{1}{7}x-\frac{2}{7}=0\)hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\)hoặc \(\frac{1}{3}x+\frac{4}{3}=0\)

+) \(\frac{1}{7}x-\frac{2}{7}=0\Leftrightarrow\frac{1}{7}x=\frac{2}{7}\Leftrightarrow x=2\)

+)\(-\frac{1}{5}x+\frac{3}{5}=0\Leftrightarrow-\frac{1}{5}x=-\frac{3}{5}\Leftrightarrow x=3\)

+)\(\frac{1}{3}x+\frac{4}{3}=0\Leftrightarrow\frac{1}{3}x=-\frac{4}{3}\Leftrightarrow x=-4\)

  c, \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{2}x=\frac{4}{9}\)

\(\Leftrightarrow x=\frac{8}{9}\)

a/ \(\frac{1}{6}x+\frac{1}{10}-\frac{4}{15}x+1=0\)

    \(\Rightarrow-\frac{1}{10}x=-\frac{11}{10}\)

     \(\Rightarrow x=11\)

b/ \(\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

   \(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{1}{7}x=\frac{2}{7}\Rightarrow x=2\)

hoặc \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow-\frac{1}{5}x=-\frac{3}{5}\Rightarrow x=3\)

hoặc \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{1}{3}x=-\frac{4}{3}\Rightarrow x=-4\)

                                              Vậy x = 2, x = 3, x = -4

c/ \(\frac{1}{2}x-\frac{11}{15}:\frac{33}{35}=-\frac{1}{3}\)

     \(\Rightarrow\frac{1}{2}x-\frac{7}{9}=-\frac{1}{3}\)

      \(\Rightarrow\frac{1}{2}x=\frac{4}{9}\Rightarrow x=\frac{8}{9}\)

                                                                       Vậy x = 8/9

Ta có : \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

<=> \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x=-1\)

<=> \(\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x=-1\)

<=> \(0.x=-1\)

=> x thuộc rỗng 

\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Leftrightarrow\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x+1=0\)

\(\Leftrightarrow\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)x+1=0\)

\(\Leftrightarrow\frac{5+3-8}{30}x+1=0\)

\(\Leftrightarrow0x=-1\)(vô lí)

vậy không có giá trị nào của x thỏa mãn. 

b, \(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

        \(0+1=0\)

=> x thuoc rong 

câu đầu nè e

x(1/6-4/15)+11/10 = 0

-x10. =-11/10

x=11

xy hình như là y/4 chứ nhỉ

\(\frac{5}{x}+\frac{4}{y}=\frac{1}{8}\)

\(\Rightarrow\frac{5}{x}=\frac{1}{8}-\frac{4}{y}\)

\(\Rightarrow\frac{5}{x}=\frac{y-32}{8y}\)

\(\text{ }\Rightarrow\orbr{\begin{cases}y-32=5\\x=8y\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}y=37\\x=8.y\end{cases}}\Rightarrow\orbr{\begin{cases}y=37\\x=8.37\end{cases}}\Rightarrow\orbr{\begin{cases}y=37\\x=296\end{cases}}\)

\(a,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(\Leftrightarrow\left[\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right]x=-1\)

\(\Leftrightarrow0x=-1\Leftrightarrow x\in\varnothing\)

\(b,\left|x\cdot\left[x^2-\frac{5}{4}\right]\right|=x\)

Vì vế trái \(\left|x\left[x^2-\frac{5}{4}\right]\right|\ge0\)với mọi x nên vế phải \(x\ge0\)

Ta có : \(x\left|x^2-\frac{5}{4}\right|=x\)vì \(x\ge0\)

Nếu x = 0 thì \(0\left|0^2-\frac{5}{4}\right|=0\)đúng

Nếu \(x\ne0\)thì ta có \(\left|x^2-\frac{5}{4}\right|=1\Leftrightarrow x^2-\frac{5}{4}=\pm1\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)

a) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

=> \(\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x+1=0\)

=> \(0x+1=0\)

=> \(1=0\)(vô lí)

b) |x . (x² - 5/4)| = x

TH1: \(x.\left(x^2-\frac{5}{4}\right)=x\)

=> \(x^3-\frac{5}{4}x-x=0\)

=> \(x^3-\frac{9}{4}x=0\)

=> \(x\left(x^2-\frac{9}{4}\right)=0\)

=> \(\orbr{\begin{cases}x=0\\x^2-\frac{9}{4}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^2=\left(\frac{3}{2}\right)^2\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm\frac{3}{2}\end{cases}}\)

TH2: \(x\left(x^2-\frac{5}{4}\right)=-x\)

=> \(x^3-\frac{5}{4}x+x=0\)

=> \(x^3-\frac{1}{4}x=0\)

=> \(x\left(x^2-\frac{1}{4}\right)=0\)

=> \(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^2=\left(\frac{1}{2}\right)^2\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=\pm\frac{1}{2}\end{cases}}\)

Do |x.(x² - 5/4)| \(\ge\)0 => x\(\ge\)0 => x thuộc {0; 1/2; 3/2}

\(a,\text{ }\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

     \(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

     \(x\left(\frac{5}{30}+\frac{3}{30}-\frac{8}{30}\right)+1=0\)

      \(x\cdot0=0-1\)

       \(0x=-1\)

\(\Rightarrow\text{ }x\in\varnothing\)

\(b,\text{ }\left|x\cdot\left(x^2-\frac{5}{4}\right)\right|=x\)

     Vì giá trị tuyệt đối của một số là một số không âm nên \(x\ge0\)

Ta có \(x\left(x^2-\frac{5}{4}\right)=\pm x\)                              \(^{\left(1\right)}\)

\(^{\text{*}}\text{ }x=0\) là một giá trị thỏa mãn                            \(^{\left(1\right)}\)

\(^{\text{*}}\)Nếu \(x=0\) thì \(^{\left(1\right)}\) \(\Leftrightarrow\text{ }x^2-\frac{5}{4}=1\pm\text{ }\Leftrightarrow\text{ }\orbr{\begin{cases}x=\pm\frac{3}{2}\\x=\pm\frac{1}{2}\end{cases}}\)

Vì có điều kiện \(x\ge0\) nên ta có đáp số \(x\in\left\{0\text{ ; }\frac{1}{2}\text{ ; }\frac{3}{2}\right\}\)