\(\frac{2}{3}\left(x-1\right)+\frac{4}{7}\left(2x-3\right)=\frac{1}{21}\left(x-6\right)\)

\(\frac{2}{3}x-\frac{2}{3}+\frac{8}{7}x-\frac{12}{7}=\frac{1}{21}x-\frac{6}{21}\)

\(\frac{2}{3}x+\frac{8}{7}x-\frac{1}{21}x=\frac{2}{3}+\frac{12}{7}-\frac{6}{21}\)

\(\frac{14}{21}x+\frac{24}{21}x-\frac{1}{21}x=\frac{14}{21}+\frac{36}{21}-\frac{6}{21}\)

\(x\left(\frac{14}{21}+\frac{24}{21}-\frac{1}{21}\right)=\frac{14+36-6}{21}\)

\(\frac{37}{21}x=\frac{44}{21}\)

\(x=\frac{44}{21}:\frac{37}{21}=\frac{44}{37}\)

cảm ơn bạn

\(=x^2-2x+4\)

= x² - 2x + 4

\(\left[x\left(x^2-2x+4\right)\right]:x=x^2-2x+4\)

a)(3x+2):8=4

3x+2=4.8

3x+2=32

3x=32-2

3x=30

x=30:3

x=10

b)x+2x+3x+4x=100

x(1+2+3+4)=100

x.10=100

x=100:10

x=10

b) x + 2x + 3x + 4x = 100

x( 1 + 2 + 3 + 4 ) = 100

x . 10 = 100

x = 100 : 10

x = 10

\(a,\text{Ta có: với mọi}\) \(x\) \(\text{thì}\) \(\left(x+2018\right)^2\ge0\)

\(\Rightarrow\orbr{\begin{cases}x+1>0;x-4< 0\\x+1< 0;x-4>0\end{cases}}\)

TH1: \(\hept{\begin{cases}x+1>0\\x-4< 0\end{cases}\text{​​}\Rightarrow\hept{\begin{cases}x>-1\\x< 4\end{cases}\Rightarrow-1< x< 4}}\)

TH2: \(\hept{\begin{cases}x+1< 0\\x-4>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>4\end{cases}\left(loại\right)}}\)

Vậy \(-1< x< 4\)

\(b.x< 2x\)

\(\Rightarrow x-2x< 0\)

\(\Rightarrow x.\left(1-2\right)< 0\)

\(-x< 0\)

\(x>0\)

\(x^3< x^2\)

\(\Rightarrow x^3-x^2< 0\)

\(\Rightarrow x^2\left(x-1\right)< 0\)

\(\Rightarrow\orbr{\begin{cases}x^2>0;\left(x-1\right)< 0\left(nhận\right)\\x^2< 0;\left(x-1\right)>0\left(loại\right)\end{cases}}\)

\(\Rightarrow x< 1\left(x\ne0\right)\)

\(4x-x^2-5< 0\)

\(=\left(-x^2-4x\right)-5=-\left(x^2-2x.2+4-4\right)-5=-\left(x-2\right)^2+4-5\)

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

Vì \(-\left(x^2-2x\right)\le0\)với mọi x nên \(-\left(x-2\right)^2-1< 0\)với mọi x 

Vậy \(4x-x^2-5< 0\)với mọi x ( đpcm ) 

4x - x² - 5 < 0 \(\forall\)x

Ta có : 4x - x² - 5 

       = -x² + 4x - 5

       = - ( x² - 4x + 5 )

       = - ( x² - 2.x.2 + 2² - 1 )

       = - [( x - 2 )² - 1 ]

Vì - ( x - 2 ) \(\le\)0 \(\forall\)x 

\(\Leftrightarrow\)- ( x - 2 ) - 1 \(\le\)0 \(\forall\)x

Vậy ..... 

4x - x² - 5 < 0 ∀x

Ta có : 4x - x² - 5 

       = -x² + 4x - 5

       = - ( x² - 4x + 5 )

       = - ( x² - 2.x.2 + 2² - 1 )

       = - [( x - 2 )² - 1 ]

Vì - ( x - 2 ) ≤0 ∀x 

⇔- ( x - 2 ) - 1 ≤0 ∀x

Vậy ..... 

P/s tham khảo nha

1,(x-1)(y+5)=101

th1:x-1=101

<=>x=102

th2:y+5=101

<=>y=96

2,(x-2)(-y+5)=12

th1:x-2=12

<=>x=14

th2:-y+5=12

<=>-y=7

<=>y=-7

3,(x-5)(2x+6)<0

th1:x-5<0

<=>x<5

th2:2x+6<0

<=>2x<-6

<=>x<-3

\(A=x^2+2x+4\)

   \(=\left(x^2+2x+1\right)+3\)

    \(=\left(x+1\right)^2+3\)

Vì \(\left(x+1\right)^2\ge0\)

\(\Rightarrow\left(x+1\right)^2+3\ge3\)

Hay \(A\ge3\)

Dấu \("="\) xảy ra khi: \(\left(x+1\right)^2=0\)

                              \(\Rightarrow x+1=0\)

                           \(\Rightarrow x=-1\) 

Vậy min A=3 đạt tại x=-1

b/ \(\left|\left|3x-1+9\right|\right|=-\left(-31\right)\)

<=> \(\left|\left|3x+8\right|\right|=31\)

<=> \(\left|3x+8\right|=31\)

<=> \(\orbr{\begin{cases}3x+8=-31\\3x+8=31\end{cases}}\)

<=> \(\orbr{\begin{cases}3x=-39\\3x=23\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-13\\x=\frac{23}{3}\end{cases}}\)