\(\frac{144}{x-3}-\frac{144}{x}=4\)

\(144x-144\left(x-3\right)=4\)

\(144x-144x+144.3=4\)

\(144x-144x+432=4\)

\(432-4\)

=> Không tìm được x thỏa mãn

nhầm 432 = 4 nhé

\(\frac{144}{x-3}-\frac{144}{x}=4\)

\(ĐKXĐ:x\ne3;x\ne0\)

\(\Leftrightarrow\frac{144x-144\left(x-3\right)}{x\left(x-3\right)}=4\)

\(\Leftrightarrow\frac{144x-144x+432}{x^2-3x}=4\)

\(\Leftrightarrow\frac{432}{x^2-3x}=4\)

\(\Leftrightarrow x^2-3x=\frac{432}{4}\)

\(\Leftrightarrow x^2-3x=108\)

\(\Leftrightarrow x^2-3x-108=0\)

\(\Leftrightarrow x^2+9x-12x-108=0\)

\(\Leftrightarrow\left(x^2+9x\right)-\left(12x+108\right)=0\)

\(\Leftrightarrow x\left(x+9\right)-12\left(x+9\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-12\right)=0\)

Hoặc \(x+9=0\Leftrightarrow x=-9\left(N\right)\)

Hoặc \(x-12=0\Leftrightarrow x=12\left(N\right)\)

Vậy tập nghiệm của phương trình là \(S=\left\{-9;12\right\}\)

\(\frac{x}{y}=\frac{3}{4}\Rightarrow\frac{x}{3}=\frac{y}{4}\) (*)

\(\frac{y}{x}=\frac{4}{5}\Rightarrow\frac{y}{4}=\frac{z}{5}\) (**)

Từ (*) và (**) có: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{3+4+5}=\frac{144}{12}=12\)

\(\frac{x}{3}=12\Rightarrow x=36\)

\(\frac{y}{4}=12\Rightarrow y=48\)

\(\frac{z}{5}=12\Rightarrow z=60\)

\(\frac{x}{y}=\frac{3}{4}\Rightarrow\frac{x}{3}=\frac{y}{4}\)

\(\frac{y}{z}=\frac{4}{5}\Rightarrow\frac{y}{4}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

áp dụng................... (bạn tự ghi nhé)

\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{x+y+z}{3+4+5}=\frac{144}{12}=12\)

tìm x,y,z biết

x/y=3/4 ; y/x=4/5

x+y+z=144 

=> x = 144 : ( 3 + 4 + 5) . 3 = 36 

=> y = 144 : ( 3 + 4 + 5 ) . 4 = 48 

=> z = 144 : ( 3 + 4 + 5 ) .5 = 60 

\(4^2.x+2^4.x+3=144\\ \Rightarrow16x+16x+3=144\\ \Rightarrow32x=144-3\\ \Rightarrow32x=141\\ \Rightarrow x=\dfrac{141}{32}\)

2x + 2x + 3 = 144

         =>  2x . 1 + 2x . 23 = 144 

          => 2x . (1 + 23) = 144

         =>  2x . 9 = 144

         => 2x = 144 : 9 = 16

        => x = 4

2x + 2x + 3 = 144

         =>  2x . 1 + 2x . 23 = 144 

          => 2x . (1 + 23) = 144

         =>  2x . 9 = 144

         => 2x = 144 : 9 = 16

        => x = 4

a: \(\Leftrightarrow x^4-24x^2+144=0\)

\(\Leftrightarrow\left(x^2-12\right)^2=0\)

hay \(x=\pm2\sqrt{3}\)

b: \(\Leftrightarrow x^4-2x^2+12x-8=0\)

\(\Leftrightarrow\left(x^2-2x+4\right)\left(x^2+2x-2\right)=0\)

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

hay \(x\in\left\{-1+\sqrt{3};-1-\sqrt{3}\right\}\)

a) \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{4}{9}\)

\(\left(x-\dfrac{1}{4}\right)^2=\left(\dfrac{2}{3}\right)^2\)

\(x-\dfrac{1}{4}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}+\dfrac{1}{4}\)

\(x=\dfrac{8}{12}+\dfrac{3}{12}\)

\(x=\dfrac{11}{12}\)

\(\)

\(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x^2-y^2\right)=144\left(1\right)\\\sqrt{x^2+y^2}-\sqrt{x^2-y^2}=y\left(2\right)\end{matrix}\right.\)

\(\Rightarrow\left(2\right)\Leftrightarrow2x^2-2\sqrt{\left(x^2+y^2\right)\left(x^2-y^2\right)}=y^2\)

\(\Leftrightarrow y^2=2x^2-24\)Thế vô (1) ta được

\(\left(x^2+2x^2-24\right)\left(x^2-2x^2+24\right)=144\)

\(\Leftrightarrow x^4-32x^2+240=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm2\sqrt{5}\\x=\pm2\sqrt{3}\end{matrix}\right.\)

Làm tiếp nhé