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\(a,=x^2+x+4x+4=\left(x+1\right)\left(x+4\right)\\ b,=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\\ c,=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\\ d,=3\left(x^2-2x+5x-10\right)=3\left(x-2\right)\left(x+5\right)\\ e,=-3x^2+6x-x+2=\left(x-2\right)\left(1-3x\right)\\ f,=x^2-x-6x+6=\left(x-1\right)\left(x-6\right)\\ h,=4\left(x^2-3x-6x+18\right)=4\left(x-3\right)\left(x-6\right)\\ i,=3\left(3x^2-3x-8x+5\right)=3\left(x-1\right)\left(3x-8\right)\\ k,=-\left(2x^2+x+4x+2\right)=-\left(2x+1\right)\left(x+2\right)\\ l,=x^2-2xy-5xy+10y^2=\left(x-2y\right)\left(x-5y\right)\\ m,=x^2-xy-2xy+2y^2=\left(x-y\right)\left(x-2y\right)\\ n,=x^2+xy-3xy-3y^2=\left(x+y\right)\left(x-3y\right)\)
Hai tam giác vuông CAB và CFE đồng dạng (chung góc C)
\(\Rightarrow\dfrac{CF}{CA}=\dfrac{EF}{AB}=\dfrac{AD}{AB}=\dfrac{AD}{3}\)
\(\Rightarrow\dfrac{AC-AF}{AC}=\dfrac{AD}{3}\Leftrightarrow\dfrac{AC-2}{AC}=\dfrac{AD}{3}\Rightarrow AD=3\left(\dfrac{AC-2}{AC}\right)\)
\(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{3}{2}AC\)
\(S_{ADEF}=AD.AF=2AD=6\left(\dfrac{AC-2}{AC}\right)\)
Theo đề bài: \(S_{ADEF}=\dfrac{1}{2}S_{ABC}\Rightarrow6\left(\dfrac{AC-2}{AC}\right)=\dfrac{1}{2}.\dfrac{3}{2}AC\)
\(\Leftrightarrow8\left(AC-2\right)=AC^2\Leftrightarrow AC^2-8AC+16=0\)
\(\Leftrightarrow\left(AC-4\right)^2=0\Leftrightarrow AC=4\)
Vậy \(S_{ABC}=\dfrac{1}{2}AB.AC=\dfrac{1}{2}.3.4=6\left(cm^2\right)\) \(\Rightarrow S_{ADEF}=3\)
a: \(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
b: \(N=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)
1. \(M=\dfrac{5}{x-1}-\dfrac{8}{x^2-1}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)
\(M=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)\(M=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\)
\(M=\dfrac{1}{x-1}.\)
2. \(N=\dfrac{5}{x-1}+\dfrac{8}{1-x^2}-\dfrac{4}{x+1}\left(x\ne\pm1\right).\)
\(N=\dfrac{5\left(x+1\right)-8-4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x+5-8-4x+4}{\left(x-1\right)\left(x+1\right)}\)
\(N=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}.\)
3. \(Q=\dfrac{1}{2x-1}-\dfrac{4}{4x^2-1}-\dfrac{2}{2x+1}\left(x\ne\pm\dfrac{1}{2}\right).\)
\(Q=\dfrac{2x+1-4-2\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x-3-4x+2}{\left(2x-1\right)\left(2x+1\right)}\)
\(Q=\dfrac{-2x-1}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-1}{2x-1}.\)
4. \(F=\dfrac{x+3}{x-2}+\dfrac{x+2}{3-x}+\dfrac{x+2}{x^2-5x+6}\left(x\ne2,x\ne3\right).\)
\(F=\dfrac{x+3}{x-2}-\dfrac{x+2}{x-3}+\dfrac{x+2}{\left(x-3\right)\left(x-2\right)}\)
\(F=\dfrac{\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x-2\right)+x+2}{\left(x-2\right)\left(x-3\right)}\)
\(F=\dfrac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(F=\dfrac{1}{x-2}.\)
xét :
|2x - 1| = 2x - 1 nếu 2x - 1 >0 hay x > \(\frac{1}{2}\)
=> 4(2x - 1) - x = 2 <=> 8x - 4 -x = 2 <=> 7x = 6 => x = \(\frac{6}{7}\)
( thỏa mãn ĐK )
|2x -1| = 1 - 2x nếu 2x - 1 < 0 hay x < \(\frac{1}{2}\)
=> 4.( 1 - 2x) - x = 2 <=> 4 - 8x -x = 2<=> 2 =9x => x = \(\frac{2}{9}\) (thỏa mãn ĐK)
vậy phương trình có nghiệm s = { \(\frac{1}{2},\frac{2}{9}\) }
\(\Leftrightarrow2\left(x+1\right)^3=56\Leftrightarrow\left(x+1\right)^3=28\Leftrightarrow\)
\(a,A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(=x-0,2-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(=\left(-0,2-2+2\right)+\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)\)
\(=-0,2\)
\(b,B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(=x^3-8y^3-x^3+8y^3-10\)
\(=-10\)
\(c,C=4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)-4x\)
\(=4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=13\)
a) \(A=0,2\left(5x-1\right)-\dfrac{1}{2}\left(\dfrac{2}{3}x+4\right)+\dfrac{2}{3}\left(3-x\right)\)
\(A=x-\dfrac{1}{5}-\dfrac{1}{3}x-2+2-\dfrac{2}{3}x\)
\(A=\left(x-\dfrac{1}{3}x-\dfrac{2}{3}x\right)-\left(\dfrac{1}{5}+2-2\right)\)
\(A=-\dfrac{1}{5}\)
Vậy: ...
b) \(B=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x^3-8y^3+10\right)\)
\(B=\left[x^3-\left(2y\right)^3\right]-\left[x^3-\left(2y\right)^3\right]-10\)
\(B=-10\)
Vậy: ...
c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)-4x\)
\(=4\left(x^2+2x+4\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-4x\)
\(=4x^2+8x+4+4x^2-4x+1-8x^2+8-4x\)
\(=\left(4x^2+4x^2-8x^2\right)+\left(8x-4x-4x\right)+\left(4+1+8\right)\)
\(=13\)
Vậy:...