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a)\(\frac{5}{9}\times x-1=\frac{8}{9}\)
\(\frac{5}{9}\times x=\frac{8}{9}+1\)
\(\frac{5}{9}\times x=\frac{17}{9}\)
\(x=\frac{17}{9}:\frac{5}{9}\)
\(x=\frac{17}{5}\)
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a) 213 - 2 x X = 47
2 x X = 213 - 47
2 x X = 166
X = 166 : 2
X = 83
b) 3 x ( X + 1 ) - 300 = 906
3 x ( X + 1 ) = 906 + 300
3 x ( X + 1 ) = 1206
X + 1 = 1206 : 3
X + 1 = 402
X = 402 - 1
X = 401
c) \(\frac{7}{3}-\frac{2}{3}\times x=\frac{1}{4}\)
\(\frac{2}{3}\times x=\frac{7}{3}-\frac{1}{4}\)
\(\frac{2}{3}\times x=\frac{25}{12}\)
\(x=\frac{25}{12}:\frac{2}{3}\)
\(x=\frac{25}{8}\)
d) \(\frac{2}{5}\times x+\frac{7}{8}=2\)
\(\frac{2}{5}\times x=2-\frac{7}{8}\)
\(\frac{2}{5}\times x=\frac{9}{8}\)
\(x=\frac{9}{8}:\frac{2}{5}\)
\(x=\frac{45}{16}\)
Học tốt #
a) \(213-2\cdot x=47\)\(\Leftrightarrow2\cdot x=166\Leftrightarrow x=83\)
b)\(3x\left(x+1\right)-300=906\Leftrightarrow3x\left(x+1\right)=606\)
\(\Leftrightarrow x\left(x+1\right)=202\Leftrightarrow x^2+x-202=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-202,25=0\)\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\sqrt{202,25}\\x+\frac{1}{2}=-\sqrt{202,25}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{202,25}-0,5\\x=-\sqrt{202,25}-0,5\end{cases}}\)
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3 x ( 5 x X - 1 ) = 10
( 5 x X - 1 ) = 10 : 3
( 5 x X - 1 ) = \(\frac{10}{3}\)
5 x X = \(\frac{10}{3}+1\)
5 x X = \(\frac{13}{3}\)
x = \(\frac{13}{3}:5\)
x = \(\frac{13}{15}\)
\(3\times\left(5\times x-1\right)=10\)
\(5\times x-1=10:3\)
\(5\times x-1=\frac{10}{3}\)
\(5\times x=\frac{10}{3}+1\)
\(5\times x=\frac{13}{3}\)
\(x=\frac{13}{3}:5\)
\(x=\frac{13}{15}\)
Học tốt #
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`@` `\text {Ans}`
`\downarrow`
`1988 \times 669 - 1988 + 332 \times 1988`
`= 1988 \times 669 - 1998 \times 1 + 332 \times 1988`
`= 1988 \times (669 - 1 + 332)`
`= 1988 \times 1000`
`= 1988000`
`@` `\text {duynam4510}`
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\(1990.x-x=1990.1988+1999\)
\(x\left(1990-1\right)=1990\left(1988+1\right)\)
\(x.1989=1990.1989\)
\(\Rightarrow x=1990\)
lưu ý: "." là nhân nha
\(1990x-x=1990\cdot1988+1990+9\)
\(x.\left(1990-1\right)=1990.\left(1988+1\right)+9\)
\(1989x=1989.1990+9\)
\(1989x-1989.1990=9\)
\(1989.\left(x-1990\right)=9\)
\(x-1990=\frac{9}{1989}\)
\(x=\frac{1}{221}+1990\)
\(x=1990\frac{1}{221}\)
Vậy \(x=1990\frac{1}{221}\)
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Ta có : 11.(x - 5) = 5x + 11
=> 11x - 55 = 5x + 11
=> 11x - 5x = 11 + 55
=> 6x = 66
=> x = 66 : 6
=> x = 11
1988 < 5 x X - 2 < 1998
1988 < [ ( 1998 - 8 ) : 5 + 1 ] x 5 - 2 < 1998
1988 < ( 1990 : 5 + 1 ) x 5 - 2 < 1998
1988 < 399 x 5 - 2 < 1998
= > Vậy số cần tìm x = 399
1988 < 5 . x - 2 < 1998
1998 < { ( 1998 - 8 ) : 5 + 1 } . 5 - 2 < 1998
1988 < ( 1990 : 5 + 1 ) . 5 - 2 < 1998
1988 < ( 398 + 1 ) . 5 - 2 < 1998
1988 < 399 . 5 - 2 < 1998
=> x = 399
Vậy ...