Xx2/3+Xx4/3=22/5
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Xx2+Xx4+Xx6+...+Xx18+Xx20
=Xx(2+4+6+...+18+20)
=Xx 110
=??????????????
( X x 2 + 3 + 4 + 1 ) = 2130
X x 10 = 2130
X = 2130 : 10
X = 213
Xx2+Xx3+Xx4+X=2130
Xx(2+3+4+1) =2130
Xx10 =2130
X =2130:10
X =213
(Xx1)+(Xx2)+(Xx3)+(Xx4)=50
=>4xX+10=50
=>4xX=50-10
=>4xX=40
=>X=40:10
=>X=4
x*2 + x*3 + x*4 + x*5 + x*6 = x*9 + 9999
=> x*(2 + 3 + 4 + 5 + 6) = x*9 + 9999
=> x*20 = x*9 + 9999
=> x*20 - x*9 = 9999
=> x*11 = 9999
=> x = 909
xx2 + xx3 +xx4+ xx5 + xx6 =xx9 + 9999
xx2 + xx3 + xx4 + xx5 + xx6 -xx9 = 9999
xx*10+2+ xx*10+3+xx*10+4+xx*10+5+xx*10+6-xx*10+9 = 9999
(xx*10+xx*10+xx*10+xx*10+xx*10-xx*10) + (2+3+4+5+6-9) = 9999
xx * (10+10+10+10+10-10) + 11 = 9999
xx*(10*5-10) = 9988
sai đề rồi
\(a,x\times2+x\times8=960\)
\(x\times\left(2+8\right)=960\)
\(x\times10=960\)
\(x=960:10\)
\(x=96\)
\(b,x\times13-x\times4=612\)
\(x\times\left(13-4\right)=612\)
\(x\times9=612\)
\(x=612:9\)
\(x=68\)
Đáp án C
f ( t ) = t ( t 2 + 3 + 1 ) ⇒ f ' ( t ) = t 2 + 3 + 1 + t t t 2 + 3 > 0 ∀ t ( x + 2 ) ( ( x + 2 ) 2 + 3 + 1 ) > − x ( x 2 + 3 + 1 ) ⇔ ( x + 2 ) ( ( x + 2 ) 2 + 3 + 1 ) > − x ( ( − x ) 2 + 3 + 1 ) ⇔ f ( x + 2 ) > f ( − x ) ⇔ x + 2 > − x ⇔ x > − 1
\(\frac{22}{5}\)\(:\)\(x\)\(=\)\(\frac{44}{5}\)\(:\)\(\frac{5}{2}\)
\(\frac{22}{5}\)\(:\)\(x\)\(=\)\(\frac{88}{25}\)
\(x\)\(=\)\(\frac{22}{5}\)\(:\)\(\frac{88}{25}\)
\(x\)\(=\)\(\frac{5}{4}\)
X * 2/3 + X * 4/3 = 22/5
x * (2/3+4/3) = 22/5
x * 6/3 = 22/5
x * 2 = 22/5
x = 22/5 : 2
x = 22/10
\(X\times\frac{2}{3}+X\times\frac{4}{3}=\frac{22}{5}\)
\(X\times\left(\frac{2}{3}+\frac{4}{3}\right)=\frac{22}{5}\)
\(X\times2=\frac{22}{5}\)
\(X=\frac{22}{5}:2\)
\(X=\frac{11}{5}\)