Tìm x , biết :
1) x-2/5 = 0,24
2) (7/3.x -0,6 ) :3 2/5 =1
giúp nha , có thưởng đó
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\(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}\) \(\frac{3}{5}.\frac{2}{8}+\frac{-6}{16}.\frac{2}{5}+\frac{-6}{15}:\left(-16\right)\)
\(=\frac{-5}{7}\left(\frac{2}{11}+\frac{9}{11}\right)\) \(=\frac{3}{20}+\frac{-3}{20}+\frac{1}{40}\)
\(=\frac{-5}{7}.1=\frac{-5}{7}\) \(=0+\frac{1}{40}=\frac{1}{40}\)
\(x-\frac{2}{5}=0,24\) \(\left(\frac{7}{3}x-0,6\right):3\frac{2}{5}=1\)
\(\Rightarrow x=0,24+\frac{2}{5}=\frac{16}{25}\) \(\Rightarrow\left(\frac{7}{3}x-0,6\right):\frac{17}{5}=1\)
vậy x = 16/25 \(\Rightarrow\frac{7}{3}x-0,6=\frac{17}{5}\)
\(\Rightarrow\frac{7}{3}x=\frac{17}{5}+0,6=4\)
\(\Rightarrow x=4:\frac{7}{3}=\frac{12}{7}\)
vậy x = 12/7
a) \(2\frac{1}{2}-\left(q-\frac{1}{2}\right)=\frac{3}{5}\)
\(\Rightarrow q-\frac{1}{2}=2\frac{1}{2}-\frac{3}{5}=\frac{19}{10}\)
\(\Rightarrow q=\frac{19}{10}+\frac{1}{2}=\frac{12}{5}=2,4\)
\(\Rightarrow q=2,4\)
b) \(\frac{1}{3}\cdot q-\frac{1}{5}\cdot q=0,6\)
\(=q\cdot\left(\frac{1}{3}-\frac{1}{5}\right)=0,6\)
\(=q\cdot\frac{2}{15}=0,6\)
\(\Rightarrow q=0,6:\frac{2}{15}=\frac{9}{2}=4,5\)
\(\Rightarrow q=4,5\)
c) \(\frac{3}{5}-q:\frac{1}{2}=\frac{1}{7}\)
\(\Rightarrow q:\frac{1}{2}=\frac{3}{5}-\frac{1}{7}=\frac{16}{35}\)
\(\Rightarrow q=\frac{16}{35}\cdot\frac{1}{2}=\frac{8}{35}\)
\(\Rightarrow q=\frac{5}{35}\)
Ở câu b áp dụng phân phối giữa phép nhân vs phép trừ.
a/ x= 3 hoặc x= -7
mới nghĩ ra có phần a thui ak thông cảm nhé
a. 9x2 - 6x - 3 = 0
<=> 3(3x2 - 2x - 1) = 0
<=> 3(3x2 - 3x + x - 1) = 0
<=> \(3\left[3x\left(x-1\right)+\left(x-1\right)\right]=0\)
<=> 3(3x + 1)(x - 1) = 0
<=> \(\left[{}\begin{matrix}3x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)
b. (2x + 1)2 - 4(x + 2)2 = 9
<=> (2x + 1)2 - \(\left[2\left(x+2\right)\right]^2=9\)
<=> (2x + 1 - 2x - 4)(2x + 1 + 2x + 4) = 9
<=> -3(4x + 5) = 9
<=> 4x + 5 = -3
<=> 5 + 3 = -4x
<=> -4x = 8
<=> -x = 2
<=> x = -2
a) \(\Leftrightarrow\left(9x^2-6x+1\right)-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2-4=0\)
\(\Leftrightarrow3\left(x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
b) \(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)
\(\Leftrightarrow12x=-24\Leftrightarrow x=-2\)
c) \(\Leftrightarrow3x^2-6x+3-3x^2+15x=21\)
\(\Leftrightarrow9x=18\Leftrightarrow x=2\)
d) \(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)
\(\Leftrightarrow2x=-40\Leftrightarrow x=-20\)
a) \(-0,6x-\dfrac{7}{3}=5,4\Leftrightarrow-\dfrac{3}{5}x=5,4+\dfrac{7}{3}\Leftrightarrow x=\dfrac{116}{15}.\left(-\dfrac{5}{3}\right)=-\dfrac{116}{9}\).
b) \(2,8:\left(\dfrac{1}{5}-3x\right)=1\dfrac{2}{5}\Leftrightarrow\dfrac{1}{5}-3x=2,8:\dfrac{7}{5}\Leftrightarrow-3x=2-\dfrac{1}{5}\Leftrightarrow x=\dfrac{9}{5}:\left(-3\right)=-\dfrac{3}{5}\).
1) x-2/5=0,24 2) (7/3 . x -0,6) : 3 . 2/5=1
x =6/25+2/5 (7/3 . x -3/5) :3 = 1: 2/5
x =6/25+10/25 (7/3 . x -3/5) :3 = 5/2
x =16/25 7/3 . x -3/5 =5/2 . 3
7/3 . x -3/5 =15/2
7/3 .x =15/2+3/5
7/3. x = 81/10
x = 81/10 : 7/3
x =243/70
2) \(\left(\frac{7}{3}.x-0,6\right):3\frac{2}{5}=1\Rightarrow\left(\frac{7}{3}x-\frac{3}{5}\right):\frac{17}{5}=1\Rightarrow\left(\frac{7}{3}x-\frac{3}{5}\right)=\frac{17}{5}\Rightarrow\frac{7}{3}x=\frac{17}{5}+\frac{3}{5}=4\Rightarrow4:\frac{7}{3}=4.\frac{3}{7}=\frac{12}{7}=1\frac{5}{7}\)