a) \(\frac38\) + x = \(\frac25\) + \(\frac14\)
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\(\frac25+100\%=\frac25+\frac{100}{100}=\frac25+1=\frac25+\frac55=\frac75\)
\(\frac25+100\%=\frac25+\frac{100}{100}=\frac{40}{100}+\frac{100}{100}=\frac{140}{100}=\frac75\)

Giải :
Gọi chiều dài là a, chiều rộng là b.
Theo đề bài ta có:
\(\frac{2}{5}\times a\times\frac{1}{4}\times b=48\)
\(\frac{1}{10}\times a\times b=48\)
\(a\times b=480\)
Vậy diện tích hình chữ nhật là 480 cm2.
~HT~
Gọi chiều dài là a, chiều rộng là b. Theo đề bài ta có:
2/5 × a × 1/4 × b = 48
1/10 × a × b = 48
a × b = 480
Vậy diện tích hình chữ nhật là 480 cm².

a) \(x+2x+3x+...+100x=-213\)
\(\Rightarrow x.\left(1+2+3+...+100\right)=-213\)
\(\Rightarrow x.5050=-213\Rightarrow x=\frac{-213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-\frac{25}{6}\)
\(\Rightarrow\frac{1}{2}x-\frac{1}{3}=\frac{-47}{12}\)
\(\Rightarrow\frac{1}{2}x=\frac{-43}{12}\Rightarrow x=\frac{-43}{6}\)
d) \(\frac{x+1}{3}=\frac{x-2}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
c) \(3\left(x-2\right)+2\left(x-1\right)=10\)
\(\Rightarrow3x-6+2x-2=10\)
\(\Rightarrow5x=18\Rightarrow x=\frac{18}{5}\)
a) \(x+2x+3x+4x+...+100x=-213\)
\(x.\left(1+2+3+4+...+100\right)=-213\)
\(x.5050=-213\)
\(x=-\frac{213}{5050}\)
b) \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{4}-4\frac{1}{6}\)
\(\frac{1}{2}x-\frac{1}{3}=-\frac{47}{12}\)
\(\frac{1}{2}x=-\frac{43}{12}\)
\(x=\frac{-43}{6}\)

\(\frac14-2x=5\)
\(2x=\frac14-5\)
\(2x=\frac{-19}{4}\)
\(x=-\frac{19}{4}:2\)
\(x=\frac{-19}{8}\)

\(P=\dfrac{\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{11}}{\dfrac{5}{12}+\dfrac{12}{12}-\dfrac{7}{11}}=\dfrac{\dfrac{5}{12}+\dfrac{5}{11}}{\dfrac{17}{12}-\dfrac{7}{11}}=\dfrac{115}{132}:\dfrac{103}{132}=\dfrac{115}{103}\)

a)P(x) = x^5 + 7x^4 - 9x^3 - 2x^2 - 1/4x
Q(x) = x^5 + 5x ^ 4 - 2x ^ 3 + 4x^2 - 1/4
b) P(x)+Q(x)
= (x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x ) + (5x^4 – x^5 + 4x^2 – 2x^3 – 1/4)
= x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x + 5x^4 – x^5 + 4x^2 – 2x^3 – 1/4
= (x^5 - x^5 ) + ( 7x^4 + 5x^4) + (-2x^3-9x^3) + ( -2x^2 +4x^2) + 1/4x+1/4
= 0 + 12x^4 + -11x^3 + 2x^2 + 1/4x + 1/4
= 12x^4 - 11x^3 + 2x^2 + 1/4x + 1/4
P(x) – Q(x)
= (x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x ) - (5x^4 – x^5 + 4x^2 – 2x^3 – 1/4)
= x^5 – 2x^2 + 7x^4 – 9x^3 – ¼ x - 5x^4 + x^5 - 4x^2 + 2x^3 + 1/4
=(x^5 + x^5 ) + ( 7x^4 - 5x^4) + (2x^3 - 9x^3) + ( -2x^2 - 4x^2) + 1/4x+1/4
= 2x^5 + 2x^4 + -7x^3 + -6x^2 + 1/4x + 1/4
=2x^5 + 2x^4 - 7x^3 - 6x^2 + 1/4x + 1/4
\(\frac38+x=\frac25+\frac14\)
\(\frac38+x=\frac{13}{20}\)
\(x=\frac{13}{20}-\frac38\)
\(x=\frac{11}{40}\)
Vậy \(x=\frac{11}{40}\)
Ta có: \(\frac38+x=\frac25+\frac14\)
=>\(x+\frac38=\frac{8}{20}+\frac{5}{20}=\frac{13}{20}\)
=>\(x=\frac{13}{20}-\frac38=\frac{26}{40}-\frac{15}{40}=\frac{11}{40}\)