\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)

\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(\Leftrightarrow x=-\frac{1}{12}\)

\(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x-1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2-2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow\)12x²+12x+3-5x²+10x-5-7x²+14x+5=0

\(\Leftrightarrow\)36x+3=0

\(\Leftrightarrow\)36x=-3

\(\Leftrightarrow\)x=\(-\frac{1}{12}\)

Vậy pt có tập nghiêm S={\(-\frac{1}{12}\)}

A B C D E K 1 2 1 2

Ta có  : \(\widehat{A_1}=\widehat{A_2}\)( do \(AD\)là phân giác )

            \(\widehat{K_1}=\widehat{K_2}\)( đối đỉnh )

Vì \(AD//KM\Rightarrow\widehat{A_2}=\widehat{K_1}\left(soletrong\right)\Rightarrow\widehat{A_1}=\widehat{K_1}\)

Mà \(\widehat{AEK}=\widehat{A_1}\)( cùng bù \(\widehat{DAE}\))

\(\Rightarrow\widehat{AEK}=\widehat{K_1}\Rightarrow\Delta AEK\)cân tại \(K\)

\(\Rightarrow AE=AK\)

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

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

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

\(\Leftrightarrow\frac{1}{4}x+\frac{3}{4}=\frac{11}{6}-\frac{5}{6}x\)

\(\Leftrightarrow\frac{1}{4}x+\frac{5}{6}x=\frac{11}{6}-\frac{3}{4}\)

\(\Leftrightarrow\frac{13}{12}x=\frac{13}{12}\)

\(\Leftrightarrow x=\frac{13}{12}\div\frac{13}{12}\)

\(\Leftrightarrow x=1\)

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

\(\Leftrightarrow\frac{1}{4}x+\frac{3}{4}=\frac{11}{6}-\frac{5x}{6}\)

\(\Leftrightarrow\frac{1}{4}x+\frac{3}{4}-\frac{3}{4}=\frac{11}{6}-\frac{5x}{6}-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}x=\frac{-5x+11}{6}-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}x.12=\frac{-5x+11}{6}.12-\frac{3}{4}.12\)

<=> 3x = 2(-5x + 11) - 9

<=> 3x = -10x + 13

<=> 3x + 10x = -10x + 13 + 10

<=> 13x = 13

=> x = 1

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

\(\Leftrightarrow\frac{4\left(4x-2\right)}{12}=\frac{24}{12}-\frac{3\left(3+5x\right)}{12}\)

\(\Leftrightarrow\frac{16x-8}{12}=\frac{24-3\left(3+5x\right)}{12}\)

\(\Leftrightarrow\frac{16x-8}{12}=\frac{24-9-15x}{12}\)

\(\Leftrightarrow16x-8=24-9-15x\)

\(\Leftrightarrow16x-8=15-15x\)

\(\Leftrightarrow16x+15x=15+8\)

\(\Leftrightarrow31x=23\)

\(\Leftrightarrow x=\frac{23}{31}\)

\(\left(4x-1\right)\left(x+3\right)-\left(2x-5\right)^2=16\)

\(\Leftrightarrow x\left(4x-1\right)+3\left(4x-1\right)-\left(2x-5\right)\left(2x-5\right)=16\)

\(\Leftrightarrow4x^2-x+12x-3-2x\left(2x-5\right)+5\left(2x-5\right)=16\)

\(\Leftrightarrow4x^2+11x-3-4x^2+10x+10x-25=16\)

\(\Leftrightarrow\left(4x^2-4x^2\right)+\left(11x+10x+10\right)-\left(3+15\right)=16\)

\(\Leftrightarrow0+31x-18=16\)

\(\Leftrightarrow31x-18=16\)

\(\Leftrightarrow31x=16+18\)

\(\Leftrightarrow31x=34\)

\(\Leftrightarrow x=\frac{34}{31}\)

\(2x^3+7x^2+7x+2=0\)

\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)

\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x^2+3x+1\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left[2x\left(x+1\right)+\left(x+1\right)\right]=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(2x+1\right)=0\)

.......................................................................................

\(x^3-8x^2-8x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)

......................................................................................

cảm ơn nha 

    B1) Tỉ số của AB=11( vì 7+4)

           Tỉ số của MA/AB=7/11

              TỈ SỐ AB/MB=   11/4

B2)          Độ dài đoạn AB= 10:2=5

               Độ dài đoạn MB =10-5

           k nhá

B1: Ta có: Tỉ số của AB là 11 ( = tỉ số MA + tỉ số MB)

=> tỉ số của MA/AB=7/11

tỉ số của AB/MB=11/4

B2: Độ dài của MA: 10/(2+3).2=4 cm

=> MB=AB-MA=10-4=6 cm

Chúc e hc tốt