1) -2001 x 25
2) 97 x (-321)
Những câu hỏi liên quan
252\x =84\97
\(\frac{252}{x}=\frac{84}{97}\)
\(\Leftrightarrow x.84=252.97\)
\(\Leftrightarrow x.84=24444\)
\(\Leftrightarrow x=\frac{24444}{84}=291\)
Vậy \(x=291\)
Chú ý: Dấu \(.\) là dấu nhân nhé
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19,96 +4,19 -24 ,15 :(:1/4-1/4)=23,15
252/x=84/97
x-2/255=114/153
*19,96+4,19-24,15:(1/4-1/4) *252/x=84/97
=19,96+4,19-24,15:0 x=252x97:84
=19,96+4,19-0 x=291
=24,15-0 * x-2/255=114/153
=24,15 x-2/255=38/51
x=38/51+2/255
x=64/85
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a ) ko rõ đề
b )\(\frac{252}{x}=\frac{84}{97}\)
=> 252 . 97 = x . 84
=> x . 84 = 24 444
x = 24 444 : 84
x = 291
Vậy x = 291
c ) x - \(\frac{2}{255}=\frac{114}{153}\)
x = 114/153 + 2/255
x = 64/85
Vậy x = 64/85
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252\ x =84\97
\(\frac{252}{x}=\frac{84}{97}\)
\(\Rightarrow252:x=\frac{84}{97}\)
\(x=252:\frac{84}{97}\)
\(x=291\)
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252 \ x =84\97
\(\frac{252}{x}\)=\(\frac{84}{97}\)
Vì 252:84=3
=> x:97=3
x=3*97
x= 291
=> \(\frac{252}{291}\)=\(\frac{84}{97}\)
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giai phuong trinh
1.
(x+1)/2004+(x+2)/2003=(x+3)/2002+(x+4)/2001
2.(201-x)/99+(203-x)/97=(205-x)/95+3
bài 2: giải các phương trình sau:
a. x -23/24 +x-23/25 = x -23/26 +x - 23/27
b. (x +2/98 +1) +(x +3/97 +1)=(x +4/96 +1) +(x +5/95 +1)
c. x+1/2004 + x+2/2003= x+3/2002 +x+4 /2001
d. 201 -x/99 + 203 -x/97 +205 -x/95 +3 =0
tim x biet
\(\frac{252}{x}\)= \(\frac{84}{97}\)
Ta có :
252/x = 84/97
=> x = 252/84/97
=> x = 291
Vậy nhé, nhớ k cho mk đó
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Bài 2: Tìm x biết:1,x^2+4x+4252,(5-2x)^2-1603,(x-3)^3-(x-3)(x^2+3x+9)+9(x+1)^2154,3(x+2)^2+(2x-1)^2-7(x-3)9x+3)365,(x-3)(x^2+3x+9)+x(x+2)(2-x)16,(2x+1)^2-4(x+2)^297,(x+3)^{^{ }2}-(x-4)(x+8)1
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Bài 2: Tìm x biết:
1,x\(^2\)+4x+4=25
2,(5-2x)\(^2\)-16=0
3,(x-3)\(^3\)-(x-3)(x\(^2\)+3x+9)+9(x+1)\(^2\)=15
4,3(x+2)\(^2\)+(2x-1)\(^2\)-7(x-3)9x+3)=36
5,(x-3)(x\(^2\)+3x+9)+x(x+2)(2-x)=1
6,(2x+1)\(^2\)-4(x+2)\(^2\)=9
7,(x+3)\(^{^{ }2}\)-(x-4)(x+8)=1
1: =>x^2+4x-21=0
=>(x+7)(x-3)=0
=>x=3 hoặc x=-7
2: =>(2x-5-4)(2x-5+4)=0
=>(2x-9)(2x-1)=0
=>x=9/2 hoặc x=1/2
3: =>x^3-9x^2+27x-27-x^3+27+9(x^2+2x+1)=15
=>-9x^2+27x+9x^2+18x+9=15
=>18x=15-9-27=-21
=>x=-7/6
6: =>4x^2+4x+1-4x^2-16x-16=9
=>-12x-15=9
=>-12x=24
=>x=-2
7: =>x^2+6x+9-x^2-4x+32=1
=>2x+41=1
=>2x=-40
=>x=-20
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Giải các phương trình sau
a) frac{x-23}{24}+frac{x-23}{25}frac{x-23}{26}+frac{x-23}{27}
b) left(frac{x+2}{98}+1right)+left(frac{x+3}{97}+1right)left(frac{x+4}{96}+1right)+left(frac{x+5}{95}+1right)
c) frac{x+1}{2004}+frac{x+2}{2003}frac{x+3}{2002}+frac{x+4}{2001}
d) frac{201-x}{99}+frac{203-x}{97}frac{205-x}{95}+30
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Giải các phương trình sau
a) \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
b) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
c) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
d) \(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3=0\)
a)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)
$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$
Do đó $x-23=0\Rightarrow x=23$
b)
PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$
$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$
$\Rightarrow x+100=0\Rightarrow x=-100$
c)
PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$
Do đó $x+2005=0\Rightarrow x=-2005$
d)
PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)
\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)
\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$
Do đó $300-x=0\Rightarrow x=300$