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1032 + X x 100 + 708 + X x 10 + 40 + X x 1 = 2000 + X
(1032 + 708 + 40) + (X x 100 + X x 10 + X x 1) = 2000 + X
1780 + X x ( 100 + 10 +1 ) = 2000 + X
1780 + X x 111 =2000 + x
1780 + X x 111 - X = 2000
1780 + X x 110 =2000
X x 110 = 2000 - 1780
X x 110 = 220
X = 220 :110
X = 2
1032+708+40+(x00+x0+x)=2000+x.
=>1780+xx0+x=2000+x.
=>xx0=2000-1780=220.
=>x=2.
Vậy x=2.
tk mk nha các bn.
-chúc ai tk mk học giỏi và may mắn-
x=0
Không tồn tại nghiệm số thực nhé bạn (88+4x=200x sai nhé )
\(1x32+7x8=200x-4x\)
\(1x32+7x8=1960\)
\(1x32=1960-7x8\) bạn thay vào rồi tính nha.
\(\Rightarrow x=2\)
1x32 +7x8 + 4x = 200x
=> x chỉ có thể là 2 hoặc 3,
x là 2 khi 7x8 bằng 728 (do 200x - 4x = 196x)
x không thể bằng 3, do 1332 + 738 khôg bằng 1960
=> x = 2
<=> 1232 + 728 + 42 = 2002
= 32x + 56x + 4x = 200x
=> x ( 32 + 56 +4 ) = 200x
=> 92 x = 200x
vì 92 khác 200 mà để 2 bên bằng nhau suy ra x phải = 0
=> x = 0
Ta có thể giải bài toán như sau:
1x32 + 7x8 + 4x = 200x => 32 + 56 + 4x = 200x => 1x + 56 + 4x = 200x => 60 + 4x = 200x => 196 = 196x => x = 1Vậy x = 1.
\(\Rightarrow1032+100xX+708+10xX+40+X=2000+X.\)
\(\Rightarrow110xX=220\Rightarrow X=220:110=2\)
a) \(\dfrac{6}{13}:\left(\dfrac{1}{2}-x\right)=\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{13}:\dfrac{15}{39}\)
\(\dfrac{1}{2}-x=\dfrac{6}{5}\)
\(x=\dfrac{1}{2}-\dfrac{6}{5}\)
\(x=-\dfrac{7}{10}\)
b) \(3\times\left(\dfrac{x}{4}+\dfrac{x}{28}+\dfrac{x}{70}+\dfrac{x}{130}\right)=\dfrac{60}{13}\)
\(3\times x\times\left(\dfrac{1}{4}+\dfrac{1}{28}+\dfrac{1}{70}+\dfrac{1}{130}\right)=\dfrac{60}{13}\)
\(x\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+\dfrac{3}{7\times10}+\dfrac{3}{7\times13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\left(1-\dfrac{1}{13}\right)=\dfrac{60}{13}\)
\(x\times\dfrac{12}{13}=\dfrac{60}{13}\)
\(x=\dfrac{60}{13}:\dfrac{12}{13}\)
\(x=5\)
\(\frac{70}{3}\left(\frac{39}{30}+\frac{39}{42}\right)-\frac{246}{7}\div\left(\frac{41}{56}+\frac{41}{72}\right)\)
\(=\frac{70}{3}\left(\frac{13}{10}+\frac{13}{14}\right)-\frac{246}{7}\div\left(\frac{41}{7\cdot8}+\frac{41}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(1+\frac{3}{10}+1-\frac{1}{14}\right)-\frac{246}{7}\div\left(\frac{40+1}{7\cdot8}+\frac{40+1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left[\left(1+1\right)+\left(\frac{3}{10}-\frac{1}{14}\right)\right]-\frac{246}{7}\div\left(\frac{5}{7}+\frac{1}{7\cdot8}+\frac{5}{9}+\frac{1}{8\cdot9}\right)\)
\(=\frac{70}{3}\left(2+\frac{8}{35}\right)-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)\right]\)
\(=\frac{70}{3}\cdot\frac{78}{35}-\frac{246}{7}\div\left[\frac{5}{7}+\frac{5}{9}+\left(\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)\right]\)
\(=\frac{35\cdot2\cdot26\cdot3}{3\cdot35}-\frac{246}{7}\div\left(\frac{5}{7}+\frac{5}{9}+\frac{1}{7}-\frac{1}{9}\right)\)
\(=52-\frac{246}{7}\div\left[\left(\frac{5}{7}+\frac{1}{7}\right)+\left(\frac{5}{9}-\frac{1}{9}\right)\right]\)
\(=52-\frac{246}{7}\div\left(\frac{6}{7}+\frac{4}{9}\right)\)
\(=52-\frac{246}{7}\div\frac{82}{63}\)
\(=52-\frac{82\cdot3\cdot9\cdot7}{7\cdot82}\)
\(=52-27=25\)
\(\frac{57}{20}-\frac{26}{15}+\frac{139}{20}\div3\)
\(=\frac{57}{20}-\frac{26}{15}+\frac{139}{60}\)
\(=\frac{171}{60}-\frac{104}{60}+\frac{139}{60}=\frac{103}{30}\)
\(\frac{39}{4}+\frac{2}{3}\left(11-\frac{23}{4}\right)\)
\(=\frac{39}{4}+11\cdot\frac{2}{3}-\frac{23}{4}\cdot\frac{2}{3}\)
\(=\frac{39}{4}+\frac{22}{3}-\frac{56}{12}\)
\(=\frac{119}{12}+\frac{88}{12}-\frac{56}{12}=\frac{151}{12}\)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2002}\right)\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2001}{2002}\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2001\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2002\cdot2003\cdot2004}=\frac{1}{2004}\)
