=>12/(x+y-1)-15/(2x-y+3)=15/2 và 12/(x+y-1)-4/(2x-y+3)=28/5

=>x+y-1=22/9; 2x-y+3=-110/19

=>x+y=31/9; 2x-y=-167/19

=>x=-914/513; y=2681/513

Có nhầm đề không bạn ?

\(\left\{{}\begin{matrix}\dfrac{4}{x+y-1}-\dfrac{5}{2x-y+3}=\dfrac{5}{2}\\\dfrac{3}{x+y-1}-\dfrac{1}{2x-y+3}=\dfrac{7}{5}\end{matrix}\right.\)

Đặt x+y-1=a; 2x-y+3=b

Theo đề, ta có: 

4/a-5/b=5/2 và 3/a-1/b=7/5

=>a=22/9; b=-110/19

=>x+y=31/9; 2x+y=-110/19-3=-167/19

=>x=-2092/171; y=2681/171

5x-2>2(x+3)\(\Leftrightarrow\)5x-2>2x+6

\(\Leftrightarrow\) 5x-2x>6+2

\(\Leftrightarrow\)3x>8

\(\Leftrightarrow\)x>\(\dfrac{8}{3}\)

Chúc bn học tốt❤

\(\left\{{}\begin{matrix}4x+3x=-6\\\dfrac{x+3y}{3}-\dfrac{y-2}{5}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7x=-6\\\dfrac{5\left(x+3y\right)-3\left(y-2\right)}{15}=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\5x+15y-3y+6=15\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\12y=9-5x=9+5\cdot\dfrac{6}{7}=9+\dfrac{30}{7}=\dfrac{93}{7}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-\dfrac{6}{7}\\y=\dfrac{93}{7\cdot12}=\dfrac{93}{84}=\dfrac{31}{28}\end{matrix}\right.\)

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\left(đkxđ:x\ne-4;-5;-6;-7\right)\)

\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-13\left(tm\right)\end{matrix}\right.\)

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\\ ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\\ \Rightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}+\dfrac{1}{\left(x^2+6x\right)+\left(7x+42\right)}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{x\left(x+4\right)+5\left(x+4\right)}+\dfrac{1}{x\left(x+5\right)+6\left(x+5\right)}+\dfrac{1}{x\left(x+6\right)+7\left(x+6\right)}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Rightarrow\dfrac{1}{x+5}-\dfrac{1}{x+7}=\dfrac{1}{18}\\ \Rightarrow\dfrac{18\left(x+7\right)}{18\left(x+5\right)\left(x+7\right)}-\dfrac{18\left(x+5\right)}{18\left(x+5\right)\left(x+7\right)}=\dfrac{\left(x+5\right)\left(x+7\right)}{18\left(x+5\right)\left(x+7\right)}\\ \Rightarrow18x+126-18x-90=x^2+5x+7x+35\\ \Leftrightarrow x^2+12x+35=36\\ \Leftrightarrow x^2+12x-1=0\\ \Leftrightarrow x^2+12x+36-37=0\\ \Leftrightarrow\left(x^2+12x+36\right)-37=0\\ \Leftrightarrow\left(x+6\right)^2-37=0\\ \Leftrightarrow\left(x+6+\sqrt{37}\right)\left(x+6-\sqrt{37}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+6+\sqrt{37}=0\\x+6-\sqrt{37}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6-\sqrt{37}\\x=\sqrt{37}-6\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{\sqrt{37}-6;-\sqrt{37}-6\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne\dfrac{\pi}{2}+k\pi\\x\ne-\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{\left(1+2cos^2x-1+2sinx.cosx\right)cosx+cos^2x-sin^2x}{1+\dfrac{sinx}{cosx}}=cosx\)

\(\Leftrightarrow\dfrac{2cos^2x\left(sinx+cosx\right)+\left(sinx+cosx\right)\left(cosx-sinx\right)}{\dfrac{sinx+cosx}{cosx}}=cosx\)

\(\Leftrightarrow\dfrac{cosx\left(sinx+cosx\right)\left(2cos^2x+cosx-sinx\right)}{sinx+cosx}=cosx\)

\(\Rightarrow2cos^2x+cosx-sinx=1\)

\(\Rightarrow cosx-sinx-cos2x=0\)

\(\Rightarrow cosx-sinx-\left(cos^2x-sin^2x\right)=0\)

\(\Rightarrow cosx-sinx-\left(cosx-sinx\right)\left(cosx+sinx\right)=0\)

\(\Rightarrow\left(cosx-sinx\right)\left(1-sinx-cosx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\) \(\Rightarrow x=\dfrac{\pi}{4}\)

Có 1 nghiệm trên khoảng đã cho

a: \(\Leftrightarrow x^3-3x^2+3x-1-x^3+2x^2-x=5x\left(2-x\right)-11\left(x+2\right)\)

=>-x^2+2x-1=10x-5x^2-11x-22

=>-x^2+2x-1=-5x^2-x-22

=>4x^2+3x+21=0

=>PTVN

b: \(\Leftrightarrow\left(x+10\right)\left(x+4\right)+3\left(x+4\right)\left(x-2\right)=4\left(x+10\right)\left(x-2\right)\)

=>x^2+14x+40+3(x^2+2x-8)=4(x^2+8x-20)

=>x^2+14x+40+3x^2+6x-24=4x^2+32x-80

=>20x+16=32x-80

=>-12x=-96

=>x=8

c: \(\Leftrightarrow6\left(x-3\right)+7\left(x-5\right)=13x+4\)

=>6x-18+7x-35=13x+4

=>-53=4(loại)

d: =>3(2x-1)-5(x-2)=3(x+7)

=>6x-3-5x+10=3x+21

=>3x+21=x+7

=>x=-7

e: =>x^3-6x^2+12x-8-x^3-3x^2-3x-1=-9x^2+1

=>-9x^2+9x-9=-9x^2+1

=>9x=10

=>x=10/9