tìm x
a) 29-x /21+ 27-x/23 + 23-x/27 + 21-x /27 =0
b) x – 5/100 + x-4/101 + x-3/102 = x-100/5 + x-101/4 +x-102/3
tìm x
a) 29-x /21+ 27-x/23 + 23-x/27 + 21-x /27 =0
b) x – 5/100 + x-4/101 + x-3/102 = x-100/5 + x-101/4 +x-102/3
Đáp án:
a, Ta có :
`(29 – x)/21 + (27 – x)/23 + (23 – x)/27 + (21 – x)/29 = -4`
`<=> (29 – x)/21 + (27 – x)/23 + (23 – x)/27 + (21 – x)/29 + 4 = 0`
`<=>[ (29 – x)/21 + 1]+ [(27 – x)/23 + 1] +[ (23 – x)/27 + 1] + [(21 – x)/29 + 1]= 0`
`<=> (50 – x)/21 + (50 – x)/23 + (50 – x)/27 + (50 – x)/29 = 0`
`<=> (50 – x)(1/21 + 1/23 + 1/27 + 1/29) = 0`
Do `1/21 + 1/23 + 1/27 + 1/29 \ne 0`
`<=> 50 – x = 0`
`<=> x = 50`
b, Ta có :
`(x – 5)/100 + (x – 4)/101 + (x- 3)/102 = (x – 100)/5 + (x – 101)/4 + (x – 102)/3`
`<=> [(x – 5)/100 – 1]+ [(x – 4)/101 – 1]+ [(x- 3)/102 – 1]= [(x – 100)/5 – 1] + [(x – 101)/4 – 1]+ [(x – 102)/3 – 1]`
`<=> (x – 105)/100 + (x – 105)/101 + (x – 105)/102 = (x – 105)/5 + (x – 105)/4 + (x – 105)/3`
`<=> (x – 105)/100 + (x – 105)/101 + (x – 105)/102 – (x – 105)/5 – (x – 105)/4 – (x – 105)/3 = 0`
`<=> (x – 105)(1/100 + 1/101 + 1/102 – 1/5 – 1/4 – 1/3) = 0`
Do `1/100 + 1/101 + 1/102 – 1/5 – 1/4 – 1/3 \ne 0`
`<=> x – 105 = 0`
`<=> x = 105`
Giải thích các bước giải:
Đáp án:
b. x=105
Giải thích các bước giải:
\(\begin{array}{l}
a.\dfrac{{29 – x}}{{21}} + \dfrac{{27 – x}}{{23}} + \dfrac{{23 – x}}{{25}} + \dfrac{{21 – x}}{{27}} = 0\\
\to \dfrac{{29 – x}}{{21}} + 1 + \dfrac{{27 – x}}{{23}} + 1 + \dfrac{{23 – x}}{{25}} + 1 + \dfrac{{21 – x}}{{27}} + 1 = 4\\
\to \dfrac{{50 – x}}{{21}} + \dfrac{{50 – x}}{{23}} + \dfrac{{50 – x}}{{25}} + \dfrac{{50 – x}}{{27}} = 0\\
\to \left( {50 – x} \right)\left( {\dfrac{1}{{21}} + \dfrac{1}{{23}} + \dfrac{1}{{25}} + \dfrac{1}{{27}}} \right) = 0\\
\to 50 – x = 0\\
\to x = 50\\
b.\dfrac{{x – 5}}{{100}} + \dfrac{{x – 4}}{{101}} + \dfrac{{x – 3}}{{102}} = \dfrac{{x – 100}}{5} + \dfrac{{x – 101}}{4} + \dfrac{{x – 102}}{3}\\
\to \dfrac{{x – 5}}{{100}} – 1 + \dfrac{{x – 4}}{{101}} – 1 + \dfrac{{x – 3}}{{102}} – 1 = \dfrac{{x – 100}}{5} – 1 + \dfrac{{x – 101}}{4} – 1 + \dfrac{{x – 102}}{3} – 1\\
\to \dfrac{{x – 105}}{{100}} + \dfrac{{x – 105}}{{101}} + \dfrac{{x – 105}}{{102}} = \dfrac{{x – 105}}{5} + \dfrac{{x – 105}}{4} + \dfrac{{x – 105}}{3}\\
\to \dfrac{{x – 105}}{{100}} + \dfrac{{x – 105}}{{101}} + \dfrac{{x – 105}}{{102}} – \dfrac{{x – 105}}{5} – \dfrac{{x – 105}}{4} – \dfrac{{x – 105}}{3} = 0\\
\to \left( {x – 105} \right)\left( {\dfrac{1}{{100}} + \dfrac{1}{{101}} + \dfrac{1}{{102}} – \dfrac{1}{5} – \dfrac{1}{4} – \dfrac{1}{3}} \right) = 0\\
\to x – 105 = 0\\
\to x = 105
\end{array}\)