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x-128=-128(cái này tự ấn máy tính nha viết hết ra lâu lắm)
x=-128+128
x=0
ko biết đúng ko nữa nếu đúng thì tick cho mình nha
x - 128 = \(\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636363}{646464}-1\right)\)
⇔ x - 128 = \(\left(\dfrac{104}{21}-5\right):\left(\dfrac{-1}{42}\right):\left(\dfrac{-1}{64}\right)\)
⇔ x - 128 = \(\left(\dfrac{-1}{21}\right):\left(\dfrac{-1}{42}\right):\left(\dfrac{-1}{64}\right)\)
⇔ x - 128 = -128
⇔ x = -128 + 128
⇔ x = 0
Vậy x = 0
\(x-128=\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636363}{646464}-1\right)\)
\(\Rightarrow x-128=\left(-\dfrac{1}{21}\right):\left(-\dfrac{1}{42}\right):\left(-\dfrac{1}{64}\right)\)
\(\Rightarrow x-128=2:\left(-\dfrac{1}{64}\right)\)
\(\Rightarrow x-128=-128\)
\(\Rightarrow x=\left(-128\right)+128\)
\(\Rightarrow x=0\)
<=> x - 128 = -25/21 : (-1/64)
<=> x - 128 = 1600/21
<=> x = 128 + 1600/21
=> x = 4288/21
Ta có : \(x-128=\left(4\frac{20}{21}-5\right)\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\\ =>x-128=\left(-\frac{1}{21}\right):\left(-\frac{1}{42}\right):\left(-\frac{1}{64}\right)\\ =>x-128=-128\\ =>x=0\)
a, Ta có x - 128 =( \(4\frac{20}{21}-5\)):\(\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\)
\(\Rightarrow\)x-128= \(\left(\frac{104}{21}-5\right):\left(\frac{41.101}{42.101}-1\right):\left(\frac{63.10101}{64.10101}-1\right)\)
\(\Rightarrow\)x-128=\(\left(\frac{-1}{21}\right):\left(\frac{-1}{42}\right):\left(\frac{-1}{64}\right)\)
\(\Rightarrow x-128=-128\)
\(\Rightarrow x=\left(-128\right)+128\)
\(\Rightarrow x=0\)
c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=0+2010\)
\(\Rightarrow x=2010\)
Vậy \(x=2010.\)
Mình chỉ làm câu c) thôi nhé.
Chúc bạn học tốt!
\(VP=\frac{1}{2\left(a+3\right)}+\frac{1}{2\left(a+5\right)}=\frac{2\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{2\left(a+3\right)\left(a+5\right)}\)
\(=\frac{2\left(a+5\right)}{4\left(a+3\right)\left(a+5\right)}+\frac{2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+5\right)+2\left(a+3\right)}{4\left(a+3\right)\left(a+5\right)}=\frac{2\left[\left(a+3\right)+\left(a+5\right)\right]}{4\left(a+3\right)\left(a+5\right)}=\frac{\left(a+3\right)+\left(a+5\right)}{2\left(a+3\right)\left(a+5\right)}\)
\(=\frac{\left(a+a\right)+\left(3+5\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{2a+8}{2\left(a+3\right)\left(a+5\right)}=\frac{2\left(a+4\right)}{2\left(a+3\right)\left(a+5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)
\(VT=\frac{x-2}{\left(a+3\right)\left(a-5\right)}\)
\(\Rightarrow\frac{x-2}{\left(a+3\right)\left(a-5\right)}=\frac{a+4}{\left(a+3\right)\left(a+5\right)}\)
\(\Rightarrow\frac{x-2}{a+4}=\frac{\left(a+3\right)\left(a-5\right)}{\left(a+3\right)\left(a+5\right)}\Rightarrow\frac{x-2}{a+4}=\frac{a-5}{a+5}\Rightarrow\left(x-2\right)\left(a+5\right)=\left(a-5\right)\left(a+4\right)\)
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