(4,5-2x) * (-1 4/7) = 11/14
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\(\left(4,5-2x\right).1\dfrac{4}{7}+\dfrac{11}{14}\)
\(=\dfrac{99}{14}-\dfrac{22}{7}x+\dfrac{11}{14}\)
\(=\left(\dfrac{99}{14}+\dfrac{11}{14}\right)-\dfrac{22}{7}x\)
\(=\dfrac{55}{7}-\dfrac{22}{7}x\)
\(=\dfrac{33}{7}+\dfrac{22}{7}.\left(1-x\right)\)
Theo bài ra, ta có:
\(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\\ \left(4,5-2x\right).\dfrac{11}{7}=\dfrac{11}{14}\\ 4,5-2x=\dfrac{11}{14}:\dfrac{11}{7}\\ 4,5-2x=\dfrac{1}{2}\\ 2x=4,5-\dfrac{1}{2}\\ \Rightarrow2x=4\\ \Rightarrow x=2\)
Vậy x = 2
\(\left(4,5-2x\right).1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\Leftrightarrow\left(4,5-2x\right).\dfrac{11}{7}=\dfrac{11}{14}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}:\dfrac{11}{7}\)
\(\Leftrightarrow4,5-2x=\dfrac{11}{14}.\dfrac{7}{11}\)
\(\Leftrightarrow4,5-2x=\dfrac{1}{2}\)
\(\Leftrightarrow-2x=\dfrac{1}{2}-4,5\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy phương trình có ngiệm x = 2 .
\(\left(4,5-2x\right)\cdot1\dfrac{4}{7}=\dfrac{11}{14}\)
\(\left(\dfrac{9}{2}-2x\right)\cdot\dfrac{11}{7}=\dfrac{11}{14}\)
\(\dfrac{9}{2}-2x=\dfrac{11}{14}:\dfrac{11}{7}\)
\(\dfrac{9}{2}-2x=\dfrac{1}{2}\)
\(2x=\dfrac{9}{2}-\dfrac{1}{2}\)
\(2x=\dfrac{8}{2}\)
\(x=\dfrac{8}{2}:2\)
`x=2`
(4,5 - 2\(x\)) (-1\(\dfrac{4}{7}\)) = \(\dfrac{11}{14}\)
(4,5 - 2\(x\)) (-\(\dfrac{11}{7}\)) = \(\dfrac{11}{14}\)
4,5 - 2\(x\) = \(\dfrac{11}{14}\) : ( - \(\dfrac{11}{7}\))
4,5 - 2\(x\) = \(\dfrac{11}{14}\) \(\times\) (- \(\dfrac{7}{11}\))
4,5 - 2\(x\) = - \(\dfrac{1}{2}\)
4,5 - 2\(x\) = -0,5
2\(x\) = 4,5 + 0,5
2\(x\) = 5
\(x\) = 5: 2
\(x\) = 2,5
(4,5-2x) * -2=11/14
9/2-2x=11/14:-2
9/2-2x=-11/28
2x=9/2 - (-11/28)
2x=137/28
x=137/28:2
x=137/56
NHA BAN
(4,5-2x) * -2=11/14
9/2-2x=11/14:-2
9/2-2x=-11/28
2x=9/2 - (-11/28)
2x=137/28
x=137/28:2
x=137/56
tao là N nè kqua phần a ra 20/3, phần b là 2.5 còn phần c tao làm rồi nhưng không chắc
(4,5-2x) * -2=11/14
9/2-2x=11/14:-2
9/2-2x=-11/28
2x=9/2 - (-11/28)
2x=137/28
x=137/28:2
x=137/56
(4,5-2x) * -2=11/14
9/2-2x=11/14:-2
9/2-2x=-11/28
2x=9/2 - (-11/28)
2x=137/28
x=137/28:2
x=137/56