Toán Lớp 8: bài 1 x ²-7=0 x ²+3x-2x-6 x ²-2x-3x-6=0 x ²-5x-24=0 bài 2 a ²-4b ² x ²+10+25 25x²-20xy+4y² 8x³+27x³ 8x mũ 6 -27x

Question

Toán Lớp 8:  bài 1 x ²-7=0 x ²+3x-2x-6 x ²-2x-3x-6=0 x ²-5x-24=0 bài 2 a ²-4b ² x ²+10+25 25x²-20xy+4y² 8x³+27x³ 8x mũ 6 -27x

ngoclankhue · Answer

Giải đáp:   $  $   B&agrave;i 1.   a,   x^2 - 7=0   -> x^2=7   ->  ( left[  begin{array}{l}x^2=( sqrt{7})^2  x^2=(- sqrt{7})^2 end{array}  right. ) ->  ( left[  begin{array}{l}x= sqrt{7}  x=- sqrt{7} end{array}  right. )    Vậy phương tr&igrave;nh c&oacute; tập nghiệm S = ( sqrt{7}, - sqrt{7})   b,   x^2 + 3x-2x-6=0   -> (x^2 +3x) - (2x+6)=0   ->x (x+3) - 2 (x+3)=0   -> (x+3) (x-2)=0   ->  ( left[  begin{array}{l}x+3=0  x-2=0 end{array}  right. ) ->  ( left[  begin{array}{l}x=-3  x=2 end{array}  right. )    Vậy phương tr&igrave;nh c&oacute; tập nghiệm S= {-3;2}   c,   x^2 - 2x-3x-6=0   -> x^2 - 5x-6=0   -> x^2 + x - 6x-6=0   -> (x^2+x) - (6x+6)=0   -> x (x+1) - 6 (x+1)=0   ->(x+1) (x-6)=0   ->  ( left[  begin{array}{l}x+1=0  x-6=0 end{array}  right. ) ->  ( left[  begin{array}{l}x=-1  x=6 end{array}  right. )    Vậy phương tr&igrave;nh c&oacute; tập nghiệm S = {-1;6}   d,   x^2 - 5x-24=0   -> x^2 - 8x + 3x-24=0   -> (x^2 - 8x)  + (3x-24)=0   -> x (x-8) + 3 (x-8)=0   -> (x-8) (x+3)=0   ->  ( left[  begin{array}{l}x-8=0  x+3=0 end{array}  right. ) ->  ( left[  begin{array}{l}x=8  x=-3 end{array}  right. )    Vậy phương tr&igrave;nh c&oacute; tập nghiệm S=  {8;-3}   $  $   B&agrave;i 2.   a,   a^2 - 4b^2   = a^2 - (2b)^2   = (a-2b) (a+2b)   b,   x^2 + 10x+25   = x^2 + 2 . 5x + 5^2   = (x+5)^2   c,   25x^2 - 20xy + 4y^2   = (5x)^2 - 2 . 5x . 2y + (2y)^2   = (5x-2y)^2   d,   8x^3 + 27y^3   = (2x)^3 + (3y)^3   = (2x+3y) (4x^2 - 6xy + 9y^2)   e,   8x^6 - 27x   = x (8x^5 - 27)

truongankimtrong · Answer

Giải đáp+Lời giải và giải thích chi tiết: Bài 1: 1)x^2-7=0 <=>x^2=7 <=> x=± sqrt{7} Vậy S={ sqrt{7};- sqrt{7}} 2)x^2+3x-2x-6=0 <=> (x^2+3x)-(2x+6)=0 <=> x(x+3)-2(x+3)=0 <=> (x+3)(x-2)=0 <=> ( left[ begin{array}{l}x+3=0 x-2=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-3 x=2 end{array} right. ) Vậy S={-3;2} 4)x^2-2x-3x-6=0 <=> (x^3+x)-(6x+6)=0 <=> x(x+1)-6(x+1)=0 <=> (x+1)(x-6)=0 <=> ( left[ begin{array}{l}x+1=0 x-6=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-1 x=6 end{array} right. ) Vậy S={-1;6} 5)x^2-5x-24=0 <=> (x^2-8x)+(3x-24)=0 <=> x(x-8)+3(x-8)=0 <=> (x-8)(x+3)=0 <=> ( left[ begin{array}{l}x-8=0 x+3=0 end{array} right. ) <=> ( left[ begin{array}{l}x=8 x=-3 end{array} right. ) Vậy S={8;-3} Bài 2: a^2-4b^2=(a-2b)(a+2b) x^2+10x+25=(x+5)^2 25x^2-20xy+4y^2=(5x-2y)^2 8x^3+27y^3=(2x+3y)(4x^2-6xy+9y^2) 8x^6-27x=x(8x^5-27)