


a)...">
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. Bài 1 : a, Ta có : \(\left(-123\right)+\left|-13\right|+\left(-7\right)\) = \(\left(-123\right)+13+\left(-7\right)=\left(-117\right)\) b, Ta có : \(\left|-10\right|+\left|45\right|+\left(-\left|-455\right|\right)+\left|-750\right|\) = \(10+45-455+750=350\) c, Ta có : \(-\left|-33\right|+\left(-15\right)+20-\left|45-40\right|-57\) = \(\left(-33\right)+\left(-15\right)+20-5-57=-90\) a) x + 15 = 36 - 2x x + 15 = 36 - (x + x ) 15 =36 - ( x + x) - x 15 = 36 - x - x - x 15 = 36 - 3x 3x = 36 - 15 3x = 21 x = 21 : 3 => x = 7 b) (x - 7) - (2x +5) = -14 x - 7 -( 2x + 5) = -14 x - (2x + 5) = -14 + 7 = -7 x - 2x - 5 = -7 x - 2x = -7 + 5 = -2 x - x + x = 2 x = 2 (-x + x cũng bằng chính nó) => x = 2 c) (x - 12) - 15 = (-7 + 20) - (18+x) (x - 12) - 15 = 13 - (18 + x) (x - 12) - 15 = 13 - 18 - x (x - 12) - 15 = -5 - x 15 = (x - 12 ) - (-5 - x) 15 = x - 12 + 5 + x 15 = x + (-12) + 5 + x 15 = 2x + [(-12) + 5] 15 = 2x + -7 2x = -7 + 15 2x = 8 x = 8 : 2 => x = 4 .................. mk sắp phải đi học rồi các bạn giúp mình với có đc ko mk nhớ sẽ đền đáp công ơn của bạn \(a)x+30\%x=-1,31\) \(\Leftrightarrow x+\frac{3x}{10}=-1,31\) \(\Leftrightarrow10x+3x=-13,1\) \(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\) \(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\) \(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\) \(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\) \(\Leftrightarrow\frac{6x-3}{2}=9\) \(\Leftrightarrow6x-3=18\) \(\Leftrightarrow x=\frac{7}{2}\) \(a,x\left(x+3\right)=0\) \(\Rightarrow\orbr{\begin{cases}x=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\) \(b,\left(x-2\right)\left(5-x\right)=0\) \(\Rightarrow\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\) \(c,\left(x-1\right)\left(x^2+1\right)=0\) \(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}\Rightarrow x=1}\) \(d,-12\left(x-5\right)+7\left(3-x\right)=15\) \(-12x+60+21-7x=15\) \(-19x+81=15\) \(-19x=15-81\) \(-19x=-66\) \(x=\frac{66}{19}\) \(e,30\left(x+2\right)-6\left(x-5\right)-24x=100\) \(30x+60-6x+30-24x=100\) \(0x+90=100\) \(0x=10\) ( vô lí ) => không có giá trị x nào thõa mãn a) x(x + 3) = 0 => \(\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\) Mà x < x + 3 => x = 0 b)( x - 2 )( 5 - x ) = 0 => \(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\5-x=0\Rightarrow x=5\end{cases}}\) => \(x\in\left\{2,5\right\}\) c) ( x - 1 )( x2 + 1 ) = 0 => \(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\x^2+1=0\Rightarrow x^2=-1\end{cases}}\) Vì x2 không thể bằng -1 => x = 1 d)-12 ( x - 5 ) + 7 ( 3 - x ) = 15 => -12x - (-60) + 21 - 7x = 15 => -12x + 60 + 21 + (-7x) = 15 =>[-12x + (-7x)] + 81 = 15 => -19x = -66 => \(x\in\varphi\) a: \(A=\dfrac{-7}{28}\cdot\dfrac{15}{25}=\dfrac{-1}{4}\cdot\dfrac{3}{5}=\dfrac{-3}{20}\) b: \(B=\dfrac{-5\cdot7}{14\cdot\left(-3\right)}=\dfrac{35}{42}=\dfrac{5}{6}\) c: \(C=\dfrac{-1}{5}-\dfrac{1}{5}\cdot\dfrac{3}{5}=\dfrac{-1}{5}-\dfrac{3}{25}=\dfrac{-8}{25}\) d: \(D=\dfrac{-3}{4}-\dfrac{1}{4}=-1\) e: \(E=\dfrac{-4}{5}\left(1-\dfrac{15}{16}\right)=\dfrac{-4}{5}\cdot\dfrac{1}{16}=\dfrac{-1}{20}\) f: \(F=\dfrac{6-7}{4}\cdot\dfrac{4+12}{22}=\dfrac{-1}{4}\cdot\dfrac{8}{11}=\dfrac{-2}{11}\) a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\) \(\Rightarrow-4< x< \dfrac{-3}{10}\) \(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\) \(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\) b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\) \(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\) \(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\) \(\Rightarrow x=\varnothing\) c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\) \(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\) \(\Rightarrow x\in\left\{1;2\right\}\) +) Với \(x=1\) \(\Rightarrow y\in\left\{1;2\right\}\) +) Với \(x=2\) \(\Rightarrow y=2\) Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).


