\(\left(x+1\right)^3-\left(x+2\right)\left(x-4\right)=\left(x-2\right)\left(x^2+2x+4\right)-2x^2\)

\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^2-2x-8\right)=x^3-8-2x^2\)

\(\Leftrightarrow3x^2+3x+1-x^2+2x+8=-8-2x^2\)

\(\Leftrightarrow4x^2+5x+17=0\)

Ta có \(\Delta=5^2-4.4.17< 0\)

Vậy pt vô nghiệm

\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)

\(4x^2-40x+100=18x+9\)

\(4x^2-58x+91=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)

\(c,x^3+3x^2-6x-8=0\)

\(\left(x+4\right)\left(x-2\right)\left(x+1\right)=0\)

\(Th1:x+4=0\Leftrightarrow x=-4\)

\(Th2:x-2=0\Leftrightarrow x=2\)

\(Th3:x+1=0\Leftrightarrow x=-1\)

\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)

\(< =>4x^2-40x+100=18x+9\)

\(< =>4x^2+58x+91=0\)

\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)

\(ĐKXĐ:x\ne2;x\ne4\)

\(\frac{x-3}{x-2}-\frac{x-2}{x-4}=\frac{16}{5}\)

\(\Leftrightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)

\(\Leftrightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)

\(\Leftrightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)

\(\Leftrightarrow16x^2-96x+128=-15x-40\)

\(\Leftrightarrow16x^2-81x+168=0\)

\(\Delta=81^2-4.16.168=-4191< 0\)

pt vô nghiệm

\(ĐKXĐ:x\ne2;x\ne4\)

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

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

\(\Leftrightarrow\frac{x^2-7x+12+x^2-4x+4}{x^2-6x+8}=-1\)

\(\Leftrightarrow\frac{2x^2-11x+16}{x^2-6x+8}=-1\)

\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)

\(\Leftrightarrow3x^2-17x+24=0\)

Ta có \(\Delta=17^2-4.3.24=1,\sqrt{\Delta}=1\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{17+1}{6}=3\\x=\frac{17-1}{6}=\frac{8}{3}\end{cases}}\)

bo deo biet

Vì a, b, c là độ dài của 3 cạnh tam giác \(\Rightarrow a,b,c>0\)

Do chu vi của tam giác bằng 1 \(\Rightarrow a+b+c=1\Rightarrow b+c=1-a\)

Giả sử : \(ab+ac+bc>a\cdot b\cdot c\)

\(\Rightarrow ab+ac+bc-abc>0\)

\(\Rightarrow a\left(b+c\right)+bc\left(1-a\right)>0\Rightarrow a\left(b+c\right)+bc\left(b+c\right)>0\)

\(\Rightarrow\left(b+c\right)\left(a+bc\right)>0\)( thỏa mãn vì \(a,b,c>0\))

Vậy \(ab+bc+ac>a\cdot b\cdot c\)( ĐPCM )

a) Áp dụng định lí Ta-lét trong \(\Delta ABC\left(DE//BC\right)\)có : 

\(\frac{AD}{AB}=\frac{DE}{BC}\Rightarrow\frac{AD}{AD+BD}=\frac{6}{16}\Rightarrow\frac{AD}{AD+10}=\frac{3}{8}\)

\(\Rightarrow8AD=3\left(AD+10\right)\Rightarrow8AD=3AD+30\Rightarrow8AD-3AD=30\)

\(\Rightarrow5AD=30\Rightarrow AD=\frac{30}{5}=6\)( cm )

b) Lấy \(F\in BC\)sao cho FC = 6cm, kẻ DF

Vì \(F\in BC\Rightarrow BF+FC=BC\)\(\Rightarrow BF+6=16\Rightarrow BF=16-6=10\)( cm )

Xét tứ giác DECF có :\(F\in BC;DE//BC\left(gt\right)\Rightarrow DE//FC\)mà \(DE=FC\left(=6cm\right)\)

\(\Rightarrow\)Tứ giác DECF là hình bình hành ( dhnb 3 ) \(\Rightarrow DF//EC\)( tính chất hình bình hành )

Hay \(DF//AC\left(E\in AC\right)\)

Áp dụng định lí Ta-lét trong \(\Delta ABC\left(DF//AC\right)\)có : 

\(\frac{BD}{AB}=\frac{BF}{BC}\)Mà lại có : \(BF=BD\left(=10cm\right)\)( cmt )

\(\Rightarrow AB=BC\Rightarrow\Delta ABC\)cân tại B ( Định nghĩa t/g cân )

** : Xin lỗi vì vẽ hình xấu nên khó nhìn, cậu hãy dùng phần chứng minh để dựng hình sao cho chuẩn nhất nhé !

a) \(7x-8=4x+7\)

\(\Leftrightarrow3x=15\)

\(\Leftrightarrow x=5\)

b) \(\frac{5x-4}{12}=\frac{16x+1}{7}\)

\(\Leftrightarrow35x-28=192x+12\)

\(\Leftrightarrow157x=-40\Leftrightarrow x=\frac{-40}{157}\)

c)\(ĐKXĐ:x\ne\pm2\)

 \(\frac{y+1}{y-2}-\frac{5}{y+2}=\frac{12}{y^2-4}+1\)

\(\Rightarrow\frac{\left(y+1\right)\left(y+2\right)-5\left(y-2\right)}{\left(y-2\right)\left(y+2\right)}=\frac{12+y^2-4}{y^2-4}\)

\(\Rightarrow\frac{y^2+3y+2-5y+10}{y^2-4}=\frac{12+y^2-4}{y^2-4}\)

\(\Rightarrow y^2-2y+12=12+y^2-4\)

\(\Rightarrow-2y=-4\Leftrightarrow y=2\left(ktm\right)\)

Vậy pt vô nghiệm

a) Thay x = 2 vào pt

\(2+4\left(2+a\right)=12\)

\(\Leftrightarrow2+8+4a=12\)

\(\Leftrightarrow4a=2\Leftrightarrow a=\frac{1}{2}\)

b) Thay x = 15 vào pt

\(\frac{a}{12}-\frac{a}{15}=3\)

\(\Leftrightarrow\frac{3a}{189}=3\Rightarrow a=189\)

a) \(2\left(x-1\right)-a\left(x-1\right)=2a+3\)

\(\Leftrightarrow2a-2-ax+a=2a+3\)

\(\Leftrightarrow-2-ax+a=3\)

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

\(\Leftrightarrow\left(x-1\right)=\frac{-5}{a}\Leftrightarrow x=\frac{a-5}{a}\)

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

\(\Leftrightarrow\frac{12x+12+8x+16+6x+18}{24}=3\)

\(\Leftrightarrow12x+12+8x+16+6x+18=72\)

\(\Leftrightarrow26x+46=72\)

\(\Leftrightarrow26x=26\Leftrightarrow x=1\)