-48+48x(-78)+48x(-21)
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-48+48.(-78)+48.(-21)
=48.(-1)+48.(-78)+48.(-21)
=48.(-1+(-78)+(-21))
=48.(-100)
=-4800
\(48\left(x-2\right)=48x+25\)
\(\Rightarrow48x-48.2=48x+25\)
\(\Rightarrow48x-96=48x+25\)
\(\Rightarrow48x-48x=25+96=121\)
\(\Rightarrow0=121\)
=> Vô lí
48 - 48 x ( 246 + 54 - 300 )
= 48 - 48 x ( 300 - 300 )
= 48 - 48 x 0
= 48 - 0
= 48
Tk Mình Nha ( MK CHUẨN BỊ LÊN LỚP 6 THUI )
\(48-48.\left(246+54-300\right)\)
\(=48-48\left(300-300\right)\)
\(=48-48.0\)
\(=48-0\)
\(=48\)
x4+4x3-4x2-48x-48=0
=> x4+4(x3-x2) - 48x = 48
=> x4 + 4[x2(x-1)] - 48x = 48
\(x^4+4x^3-4x^2-48x-48=0\)
\(\Leftrightarrow\)\(x^4-2x^3-4x^2+6x^3-12x^2-24x+12x^2-24x-48=0\)
\(\Leftrightarrow\)\(x^2\left(x^2-2x-4\right)+6x\left(x^2-2x-4\right)+12\left(x^2-2x-4\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-2x-4\right)\left(x^2+6x+12\right)\)
\(\Leftrightarrow\)\(\left[\left(x-1\right)^2-5\right]\left(x^2+6x+12\right)=0\)
\(\Leftrightarrow\)\(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\left(x^2+6x+12\right)=0\)
Ta có: \(x^2+6x+12=\left(x+3\right)^2+3>0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-1-\sqrt{5}=0\\x-1+\sqrt{5}=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}\)
Vậy...
\(\Leftrightarrow\dfrac{3x-1}{\left(6x-7\right)\left(3x+4\right)}-\dfrac{4x}{\left(8x-3\right)\left(3x+4\right)}=\dfrac{3}{\left(8x-3\right)\left(6x-7\right)}\)
=>(3x-1)(8x-3)-4x(6x-7)=3(3x+4)
=>24x^2-9x-8x+3-24x^2+28x=9x+12
=>11x+3=9x+12
=>2x=9
=>x=9/2
\(48x:\frac{21}{4}=7.5:\frac{25}{8}\)
\(48x:\frac{21}{4}=\frac{12}{5}\)
\(48x=\frac{12}{5}\cdot\frac{21}{4}\)
\(48x=\frac{63}{5}\)
\(x=\frac{63}{5}:48\)
\(x=\frac{21}{80}\)
\(3x^3-48x=0\)
\(3x\cdot\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\left\{\pm4\right\}\end{cases}}\)
Vậy,............
-48+[48x(-78+-21)]
=-48+48x-99
=48x(-99+-1)
=48x-100
=-4800
- 48 + 48 . ( - 78 ) + 48 . ( - 21 )
= 48. [ -1 + ( - 78 ) + ( - 21 ) ]
= 48. ( - 100 )
= - 4800