Bo thi:>

+ đk x > 0 , x khác 1

Mọi người ơi, giúp em với ạ!

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\) 

\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)

Đề sai chắc chỗ cuối là \(\dfrac{17}{72}\)

\(S=\dfrac{3}{2}-\dfrac{5}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}+\dfrac{15}{56}-\dfrac{17}{72}=\dfrac{1+2}{1.2}-\dfrac{2+3}{2\cdot3}+\dfrac{3+4}{3.4}-\dfrac{4+5}{4\cdot5}+\dfrac{5+6}{5.6}-\dfrac{6+7}{6.7}+\dfrac{7+8}{7.8}-\dfrac{8+9}{8\cdot9}=\dfrac{1}{2}+1-\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{5}-\dfrac{1}{7}-\dfrac{1}{6}+\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{8}=1-\dfrac{1}{9}=\dfrac{8}{9}\)

\(Đk:\) \(x\ne1,x\ne2,x\ne3\)

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

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

\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Rightarrow0x-14=x-10\)

\(\Rightarrow x=-4\left(tmđk\right)\)

b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{y+z+1}{x}=\dfrac{x+z+2}{y}=\dfrac{x+y-3}{z}=\dfrac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)

=>y+z+1=2x; x+z+2=2y; x+y-3=2z

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y-z=1\\x-2y+z=-2\\x+y-2z=3\end{matrix}\right.\Leftrightarrow\left(x,y,z\right)\in\varnothing\)

a: 3x=2y nên x/3=y/2

7x=5z nên x/5=z/7

=>x/15=y/10=z/21

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{21}=\dfrac{x-y+z}{15-10+21}=\dfrac{32}{26}=\dfrac{16}{13}\)

Do đó: x=240/13; y=160/13; z=336/13

Lời giải:

HPT \(\Leftrightarrow \left\{\begin{matrix} \frac{12}{x-1}-\frac{20}{y+4}=16\\ \frac{12}{x-1}-\frac{3}{y+1}=\frac{57}{5}\end{matrix}\right.\)

\(\Rightarrow \frac{-20}{y+4}+\frac{3}{y+1}=\frac{23}{5}\)

\(\Leftrightarrow \frac{3(y+4)-20(y+1)}{(y+1)(y+4)}=\frac{23}{5}\)

\(\Leftrightarrow \frac{-8-17y}{(y+1)(y+4)}=\frac{23}{5}\)

\(\Rightarrow 23(y+1)(y+4)+5(17y+8)=0\)

\(\Leftrightarrow 23y^2+250y+132=0\)

\(\Rightarrow y=\frac{-125\pm \sqrt{12589}}{23}\). Thay vào tìm $x$

P/s: Có vẻ bạn viết sai đề, chứ số quá xấu.

\(\dfrac{7}{9}< \dfrac{x}{y}< \dfrac{7}{8}\)

\(\dfrac{56}{72}< \dfrac{x}{y}< \dfrac{63}{72}\)

\(\dfrac{x}{y}=\left\{\dfrac{57}{72};\dfrac{58}{72};\dfrac{59}{72};\dfrac{60}{72};\dfrac{61}{72};\dfrac{62}{72}\right\}\)