a, 12 - (2\(x^2\) - 3) = 7

            2\(x^2\)  - 3  =  12  - 7

           2\(x^2\) - 3  = 5

           2\(x^2\)  = 8

             \(x^2\)   = 4

             \(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)

a) \(12-\left(2x^2-3\right)=7\\ 12-2x^2+3=7\\ 15-2x^2=7\\ 2x^2=15-7=8\\ x^2=8:2=4\\ x=\pm2\)

b) \(3x^2-12=2x^2+4\\ 3x^2-2x^2=12+4\\ x^2=16\\ x=\pm4\)

b, 3\(x^2\) - 12  = 2\(x^2\) + 4

    3\(x^2\) - 2\(x^2\) = 12 + 4

     \(x^2\) = 16

      \(\left[{}\begin{matrix}x=-4\\x=4\end{matrix}\right.\)

1.

a, => 21-x+3 < 0

=> 24-x < 0

=> x < 24

b, => 7+x > 0

=> x > -7

c, => x-1 < 0 ; x+2 > 0 ( vì x-1 < x+2 )

=> x < 1 ; x > -2

=> -2 < x < 1

Tk mk nha

3x(x² - 4) = 0

Mà 3 khác 0 

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2;2\end{cases}}\)

a, 11/12 - ( 2/5 + x ) = 2/3

<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)

b, 2x . ( x - 1/7 ) = 0

<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

vậy x={\(0;\frac{1}{7}\)}

c, 3/4 + 1/4 : x = 2/5

<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)

vậy x=-5/7

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)

\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)

\(\Leftrightarrow x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)

\(\Leftrightarrow4x=\frac{-20}{7}\)

\(\Leftrightarrow x=-\frac{5}{7}\)

a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}\)

\(\frac{2}{5}+x=\frac{1}{4}\)

\(x=\frac{1}{4}-\frac{2}{5}\)

\(x=-\frac{3}{20}\)

Vậy \(x=-\frac{3}{20}\)

b) \(2x\left(x-\frac{1}{7}\right)=0\)

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

Vậy x = 0 hoặc \(x=\frac{1}{7}\)

c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

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

\(\frac{1}{4}:x=-\frac{7}{20}\)

\(x=\frac{1}{4}:\left(-\frac{7}{20}\right)\)

\(x=-\frac{5}{7}\)

Vậy \(x=-\frac{5}{7}\)

1 là đăng lại ảnh 

2 là tách nhỏ ra

II . Hình học 

Bài 2 :

BC = AB - AC = 10 - 8 = 2 (cm )

Vì C là trung điểm của BN

nên CN = CB = 2 (cm)

NB = NC + BC = 2+ 2 = 4( cm )

Vậy CN = 2m và NB = 4 cm

có cái khác rồi

a; \(x\) - \(\dfrac{3}{5}\) = 1 - \(\dfrac{4}{5}\) + \(\dfrac{1}{6}\)

    \(x\) - \(\dfrac{3}{5}\) = \(\dfrac{30}{30}\) - \(\dfrac{24}{30}\) + \(\dfrac{5}{30}\)

    \(x\) - \(\dfrac{3}{5}\) = \(\dfrac{6}{30}\) + \(\dfrac{5}{30}\)

    \(x\) - \(\dfrac{3}{5}\) =  \(\dfrac{11}{30}\)

   \(x\)        = \(\dfrac{11}{30}\) + \(\dfrac{3}{5}\)

   \(x\)        = \(\dfrac{11}{30}\) + \(\dfrac{18}{30}\)

    \(x\)      = \(\dfrac{29}{30}\)

Vậy \(x\) = \(\dfrac{29}{30}\) 

b; (- \(\dfrac{10}{4}\)) + \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) thế \(x\) của em đâu nhỉ???

c; - \(\dfrac{3}{2}\) + (\(x\) - \(\dfrac{1}{2}\)) = \(\dfrac{1}{2}\)

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

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

             \(x\)        = 2 + \(\dfrac{1}{2}\)

             \(x\)       =   \(\dfrac{4}{2}\) + \(\dfrac{1}{2}\)

             \(x\)       = \(\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

d; \(x\) - \(\dfrac{2}{7}\) = \(\dfrac{7}{21}\)

    \(x\)       = \(\dfrac{7}{21}\) + \(\dfrac{2}{7}\)

     \(x\)     =  \(\dfrac{7}{21}\) + \(\dfrac{6}{21}\)

     \(x\)     = \(\dfrac{13}{21}\)

Vậy \(x=\dfrac{13}{21}\)