\(\left(5x-29\right)-\left(2x-29\right)=-21\)

\(5x-29-2x+29=-21\)

\(5x-2x=-21+29-29\)

\(3x=-21\)

\(x=-7\)

=>(y+5)(2x-3)=29

=>\(\left(2x-3;y+5\right)\in\left\{\left(1;29\right);\left(29;1\right);\left(-1;-29\right);\left(-29;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;24\right);\left(16;-4\right);\left(1;-34\right);\left(-13;-6\right)\right\}\)

Bạn nhấn vô biểu tượng\(\Sigma\) ở bên trái mà soạn chứ đăng bài kiểu này khó hiểu qá!

\(\sqrt{\left(x-1\right)^2+4}+\sqrt{\left(x+1\right)^2+9}=\sqrt{29}\)

\(the,a=\left(x-1\right)^2+4\)

\(\sqrt{a}+\sqrt{a+5}=\sqrt{29}\)

\(a+a+5+2\sqrt{a^2+5a}=29\)

\(2a+2\sqrt{a^2+5a}=24\)

\(a+\sqrt{a^2+5a}=12\)

\(\sqrt{a^2+5a}=12-a\)

\(a^2+5a=144-24a+a^2\)

\(29a=144\)

\(a=\frac{144}{29}\)

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

hay x=-2

\(a.DKXD:x\ne5\\ \frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow12x-9=29x-145\\ \Leftrightarrow12x-29x=9-145\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=8\)

Vậy nghiệm của phương trình trên là \(8\)

\(b.DKXD:x\ne\frac{5}{3}\\ \frac{2x-1}{5-3x}=2\\ \Leftrightarrow2x-1=2\left(5-3x\right)\\ \Leftrightarrow2x-1=10-6x\\ \Leftrightarrow2x+6x=1+10\\ \Leftrightarrow8x=11\\ \Leftrightarrow x=\frac{11}{8}\)

Vậy nghiệm của phương trình trên là \(\frac{11}{8}\)

\(c.DKXD:x\ne1\\ \frac{4x-5}{x-1}=2+\frac{x}{x-1}\\ \Leftrightarrow\frac{4x-5}{x-1}=\frac{2\left(x-1\right)+x}{x-1}\\ \Leftrightarrow4x-5=2x-2+x\\ \Leftrightarrow4x-2x-x=5-2\\ \Leftrightarrow x=3\left(tmdk\right)\)

Vậy nghiệm của phương trình trên là \(3\)

\(d.DKXD:x\ne5;-2\\ \frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7x-35=3x+6\\ \Leftrightarrow7x-3x=35+6\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)

Vậy nghiệm của phương trình trên là \(\frac{41}{4}\)

\(e.DKXD:x\ne0;-5\\\Leftrightarrow \frac{2x+5}{2x}-\frac{x}{x+5}=0\\\Leftrightarrow \frac{2x+5}{2x}=\frac{x}{x+5}\\ \Leftrightarrow\left(x+5\right)\left(2x+5\right)=2x.x\\\Leftrightarrow 2x^2+5x+10x+25=2x^2\\ \Leftrightarrow2x^2-2x^2+15x=-25\\ \Leftrightarrow15x=-25\\\Leftrightarrow x=-\frac{5}{3}\)

Vậy nghiệm của phương trình trên là \(-\frac{5}{3}\)

a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)

\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)

\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)

\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)

b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)

\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)

\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)

\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)

a: ta có: \(7-\left(2x-\frac13\right)^2=3\)

=>\(\left(2x-\frac13\right)^2=7-3=4\)

=>\(\left[\begin{array}{l}2x-\frac13=2\\ 2x-\frac13=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=2+\frac13=\frac73\\ 2x=-2+\frac13=-\frac53\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac73:2=\frac76\\ x=-\frac53:2=-\frac56\end{array}\right.\)

b: \(\left(2x+\frac13\right)^2-\frac38=\frac18\)

=>\(\left(2x+\frac13\right)^2=\frac38+\frac18=\frac48=\frac12\)

=>\(2x+\frac13=\sqrt{\frac12}=\frac{\sqrt2}{2}\)

=>\(2x=\frac{\sqrt2}{2}-\frac13=\frac{3\sqrt2-2}{6}\)

=>\(x=\frac{3\sqrt2-2}{12}\)

c: \(12:\left\lbrack29-\left(x-\frac23\right)^2\right\rbrack=3\)

=>\(29-\left(x-\frac23\right)^2=12:3=4\)

=>\(\left(x-\frac23\right)^2=29-4=25\)

=>\(\left[\begin{array}{l}x-\frac23=5\\ x-\frac23=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5+\frac23=\frac{17}{3}\\ x=-5+\frac23=-\frac{13}{3}\end{array}\right.\)

d: \(\left(3x-\frac12\right)^3+\frac83=\frac{29}{9}-\frac{14}{27}\)

=>\(\left(3x-\frac12\right)^3+\frac83=\frac{87}{27}-\frac{14}{27}=\frac{73}{27}\)

=>\(\left(3x-\frac12\right)^3=\frac{73}{27}-\frac83=\frac{73}{27}-\frac{72}{27}=\frac{1}{27}=\left(\frac13\right)^3\)

=>\(3x-\frac12=\frac13\)

=>\(3x=\frac12+\frac13=\frac56\)

=>\(x=\frac{5}{18}\)

e: \(2\left(2x-\frac13\right)^2+\frac43=\frac56+\frac{13}{18}\)

=>\(2\left(2x-\frac13\right)^2=\frac{15}{18}+\frac{13}{18}-\frac43=\frac{28}{18}-\frac{24}{18}=\frac{4}{18}=\frac29\)

=>\(\left(2x-\frac13\right)^2=\frac19\)

=>\(\left[\begin{array}{l}2x-\frac13=\frac13\\ 2x-\frac13=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac23\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=0\end{array}\right.\)

=>5.(25.2-29)-2.5.(29-5)

=>5.21-10.24

=>105-240

=-135

135 đó