\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}+2\right)+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}+2\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow10\left(x^2+\dfrac{1}{x^2}\right)+20+5\left(x^2+\dfrac{1}{x^2}\right)^2-5\left(x^2+\dfrac{1}{x^2}\right)^2-10\left(x^2+\dfrac{1}{x^2}\right)=\left(x-5\right)^2-5\)

\(\Leftrightarrow\left(x-5\right)^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=10\end{matrix}\right.\)

a: \(\Leftrightarrow x^2+x-6+2x-6=10x-20+50\)

\(\Leftrightarrow x^2+3x-12-10x-30=0\)

\(\Leftrightarrow x^2-7x-42=0\)

\(\text{Δ}=\left(-7\right)^2-4\cdot1\cdot\left(-42\right)=217>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{217}}{2}\\x_2=\dfrac{7+\sqrt{217}}{2}\end{matrix}\right.\)

b: \(\Leftrightarrow x^2-3x+5=-x^2+4\)

\(\Leftrightarrow2x^2-3x+1=0\)

\(\Leftrightarrow\left(2x-1\right)\left(x-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{2};1\right\}\)

\(\Leftrightarrow x-1-4=5\left(x-5\right)\)

=>x-5=5(x-5)

=>x-5-5x+25=0

=>-4x+20=0

hay x=5(loại)

`1/[x-5]-4/[(x-5)(x-1)]=5/[x-1]`       `ĐK: x \ne 5,x \ne 1`

`<=>[x-1-4]/[(x-5)(x-1)]=[5(x-5)]/[(x-5)(x-1)]`

   `=>x-5=5x-25`

`<=>4x=20`

`<=>x=5` (ko t/m)

Vậy ptr vô nghiệm

\(ĐK:x\ne2;x\ne-3\\ PT\Leftrightarrow\left(x-2\right)\left(x+3\right)+2\left(x+3\right)=10\left(x-2\right)+50\\ \Leftrightarrow x^2+x-6+2x+6=10x-20+50\\ \Leftrightarrow x^2-13x-30=0\\ \Leftrightarrow x^2-15x+2x-30=0\\ \Leftrightarrow\left(x-15\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\x=-2\end{matrix}\right.\left(tm\right)\)

Câu 1:

Ta có: \(\dfrac{x-4}{y-3}=\dfrac{4}{3}\)

=> \(3.\left(x-4\right)=4.\left(y-3\right)\)

=>\(3x-12=4y-12\)

=>\(3x=4y\) (1)

Ta có: \(x-y=5\)

=> \(y=y+5\) Thay vào (1) ta có:

\(3.\left(y+5\right)=4.\)y

=>\(3y+15=4y\)

=> \(15=4y-3y\)

=> 15 = y

=> y =15

ta có: x = y +5

=> x = 15 +5

=> x =20

Câu 2:

\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+...+\dfrac{10}{1400}\)

\(B=\dfrac{5}{28}+\dfrac{6}{70}+\dfrac{5}{130}+...+\dfrac{5}{700}\)

\(B=\dfrac{5}{4.7}+\dfrac{5}{7.10}+\dfrac{5}{10.13}+...+\dfrac{5}{25.28}\)

\(B=5,\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(3B=5.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{25.28}\right)\)

\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{25}-\dfrac{1}{28}\right)\)

\(3B=5.\left(\dfrac{1}{4}-\dfrac{1}{28}\right)\)

\(3B=5.\dfrac{3}{14}\)

\(B=\dfrac{15}{14}:3=\dfrac{5}{14}\)

Câu 3:

38 - (|x+10|+13) = \(\left(-6\right)^{20}:\left(9^9.4^{10}\right)\)

=> \(38-\left(\left|x+10\right|+13\right)=\left(2.3\right)_{ }^{20}:\)\(\left[\left(3^2\right)^9.\left(2^2\right)^4\right]\)

=>\(38-\left(\left|x+10\right|+13\right)=2^{20}.3^{20}:\left(3^{18}.2^{20}\right)\)

=> \(38-\left(\left|x+10\right|+13\right)=\dfrac{3^{20}.2^{20}}{3^{18}.2^{20}}\)

=> \(38-\left(\left|x+10\right|+13\right)=9\)

=> |x +10| + 13 = 38 -9

=> |x+10| +13 = 29

=> |x+10| = 29 -13

=> |x+10| = 16

\(\Rightarrow\left[{}\begin{matrix}x+10=16\\x+10=-16\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-26\end{matrix}\right.\)

Câu 2:

\(B=\dfrac{10}{56}+\dfrac{10}{140}+\dfrac{10}{260}+....+\dfrac{10}{1400}\)
\(\Rightarrow B=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+.....+\dfrac{20}{2800}\)
\(\Rightarrow B=20\left(\dfrac{1}{112}+\dfrac{1}{280}+\dfrac{1}{520}+...+\dfrac{1}{2800}\right)\)
\(\Rightarrow B=20\left(\dfrac{1}{8.14}+\dfrac{1}{14.20}+\dfrac{1}{20.26}+...+\dfrac{1}{50.56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+...+\dfrac{6}{50.56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+...+\dfrac{1}{50}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{1}{8}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20}{6}\left(\dfrac{7}{56}-\dfrac{1}{56}\right)\)
\(\Rightarrow B=\dfrac{20.6}{6.56}\)
\(\Rightarrow B=\dfrac{20}{56}\)
\(\Rightarrow B=\dfrac{5}{14}\)

\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-7}{93}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)=>x-100=0

hay x=100

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)