tìm x
\(\dfrac{x-11}{101}-\dfrac{x-13}{103}=\dfrac{x-17}{107}-\dfrac{x-19}{109}\)
DA
Những câu hỏi liên quan
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\)
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\\ \Leftrightarrow\dfrac{315-x}{101}+1+\dfrac{313-x}{103}+1+\dfrac{311-x}{105}+1+\dfrac{309-x}{107}+1=0\\ \Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\\ \Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\\ \dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}>0\\ \Rightarrow416-x=0\\ \Leftrightarrow x=416\)
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giải các phương trình sau:
1) \(\dfrac{x-11}{111}+\dfrac{x-12}{112}=\dfrac{x-23}{123}+\dfrac{x-24}{124}\)
2) \(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1980}{15}+\dfrac{x-1990}{5}\)
3) \(\dfrac{109-x}{91}+\dfrac{107-x}{93}+\dfrac{105-x}{95}+\dfrac{103-x}{97}=-4\)
Tìm x biết:
\(\dfrac{313-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=-4\)
Đề sai sửa lại và làm:
Ta có:
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{315-x}{101}+1\right)+\left(\dfrac{313-x}{103}+1\right)+\left(\dfrac{311-x}{105}+1\right)+\left(\dfrac{309-x}{107}+1\right)=0\)
\(\Leftrightarrow\dfrac{416-x}{101}+\dfrac{416-x}{103}+\dfrac{416-x}{105}+\dfrac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\dfrac{1}{101}+\dfrac{1}{103}+\dfrac{1}{105}+\dfrac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
VẬY....
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Tìm x,biết:
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=0\)
Lời giải:
PT \(\Leftrightarrow \frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=4\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=4\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=4\)
\(\Rightarrow 416-x=\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
\(\Rightarrow x=416-\frac{4}{\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}}\)
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Tìm x, biết
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}=4\)
Lời giải:
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}=4\)
\(\Leftrightarrow \frac{315-x}{101}-1+\frac{313-x}{103}-1+\frac{311-x}{105}-1+\frac{309-x}{107}-1=0\)
\(\Leftrightarrow \frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow (416-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra $416-x=0$
\(\Rightarrow x=416\)
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Giải Phương trình
1)\(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=4\)
2)\(\dfrac{x+14}{86}+\dfrac{x+15}{85}+\dfrac{x+16}{84}+\dfrac{x+17}{83}+\dfrac{x+116}{4}=0\)
3)\(x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)
Bài 1:
\(\frac{99-x}{101}+\frac{97-x}{103}+\frac{95-x}{105}+\frac{93-x}{107}=-4\)
\(\Leftrightarrow \frac{99-x}{101}+1+\frac{97-x}{103}+1+\frac{95-x}{105}+1+\frac{93-x}{107}+1=0\)
\(\Leftrightarrow \frac{99-x+101}{101}+\frac{97-x+103}{103}+\frac{95-x+105}{105}+\frac{93-x+107}{107}=0\)
\(\Leftrightarrow \frac{200-x}{101}+\frac{200-x}{103}+\frac{200-x}{105}+\frac{200-x}{107}=0\)
\(\Leftrightarrow (200-x)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
Vì \(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\neq 0\) nên suy ra \(200-x=0\Rightarrow x=200\)
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Bài 2:
\(\frac{x+14}{86}+\frac{x+15}{85}+\frac{x+16}{84}+\frac{x+116}{4}=0\)
\(\Leftrightarrow \frac{x+14}{86}+1+\frac{x+15}{85}+1+\frac{x+16}{84}+1+\frac{x+17}{83}+1+\frac{x+116}{4}-4=0\)
\(\Leftrightarrow \frac{x+100}{86}+\frac{x+100}{85}+\frac{x+100}{84}+\frac{x+100}{83}+\frac{x+100}{4}=0\)
\(\Leftrightarrow (x+100)\left(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\right)=0\)
Vì \(\frac{1}{86}+\frac{1}{85}+\frac{1}{84}+\frac{1}{83}+\frac{1}{4}\neq 0\). Do đó \(x+100=0\Rightarrow x=-100\)
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1/
99-x/101+97-x/103+95-x/105+93-x/107
\(\Leftrightarrow\)(99-x/101+1)+(97-x/103+1)+(95-x/105+1)+(93-x/107+1)=0
\(\Leftrightarrow\)100-x/101+100-x/103+100-x/105+100-x/107=0
\(\Leftrightarrow\)(100-x)(1/101+1/103+1/105+1/107)=0
\(\Leftrightarrow\)100-x=0\(\Leftrightarrow\)
x=-100
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a) Tìm x biếtdfrac{315-x}{101}+dfrac{313-x}{103}+dfrac{311-x}{105}+dfrac{309-x}{107}+40b) Cho a,b,c là các số thực khác 0 thỏa mãn dfrac{a-b+c}{b}dfrac{a+b-c}{c}dfrac{-a+b+c}{a}Tính giá trị của biểu thức :Pdfrac{left(a+bright)left(b+cright)left(c+aright)}{abc}
Đọc tiếp
a) Tìm x biết
\(\dfrac{315-x}{101}+\dfrac{313-x}{103}+\dfrac{311-x}{105}+\dfrac{309-x}{107}+4=0\)
b) Cho a,b,c là các số thực khác 0 thỏa mãn
\(\dfrac{a-b+c}{b}=\dfrac{a+b-c}{c}=\dfrac{-a+b+c}{a}\)
Tính giá trị của biểu thức :
P=\(\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Tính nhanh:
\(\dfrac{17}{13}:\dfrac{51}{26}=\dfrac{17}{13}x\dfrac{26}{51}=\dfrac{17x13x2}{13x17x3}=\dfrac{2}{3}\)
\(\dfrac{33}{38}:\dfrac{11}{19}\)=?
\(\dfrac{33}{38}:\dfrac{11}{19}=\dfrac{33\times19}{38\times11}=\dfrac{3\times1}{2\times1}=\dfrac{3}{2}\)
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\(\dfrac{2009-x}{7}+\dfrac{2007-x}{9}+\dfrac{2005-x}{11}+\dfrac{2003-x}{13}=\dfrac{x-17}{-1999}+\dfrac{x-15}{-2001}+\dfrac{x-13}{-2003}+\dfrac{x-11}{-2005}\)