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Bài làm:
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)
\(=\left(x^2+5x+5\right)^2\)
b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)
c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)
\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Làm mẫu cho 1 vd:
a, (x+1)(x+2)(x+3)(x+4)+1
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(1)
Đặt \(y=x^2+5x+5\)
Khi đó ::
(1) = \(\left(y-1\right)\left(y+1\right)+1\)
\(=y^2-1+1=y^2\)
Thay vào ta được: \(\left(x^2+5x+5\right)^2\)
a) \( - 2{x^2} + 6{x^2} = ( - 2 + 6).{x^2} = 4{x^2}\);
b) \(4{x^3} - 8{x^3} = (4 - 8).{x^3} = - 4{x^3}\);
c) \(3{x^4}( - 6{x^2}) = 3.( - 6).{x^4}.{x^2} = - 18{x^{4 + 2}} = - 18{x^6}\);
d) \(( - 24{x^6}):( - 4{x^3}) = ( - 24: - 4).({x^6}:{x^3}) = 6{x^{6 - 3}} = 6{x^3}\).
a) \(\begin{array}{l}(8{x^3} + 2{x^2} - 6x):(4x) = 8{x^3}:(4x) + 2{x^2}:(4x) - (6x):(4x)\\ = (8:4).({x^3}:x) + (2:4).({x^2}:x) - (6:4).(x:x)\\ = 2{x^2} + \dfrac{1}{2}x - \dfrac{3}{2}\end{array}\)
b) \(\begin{array}{l}(5{x^3} - 4x):( - 2x) = 5{x^3}:( - 2x) - 4x:( - 2x) = (5: - 2).({x^3}:x) - (4: - 2).(x:x)\\ = - \dfrac{5}{2}{x^{3 - 1}} - ( - 2) = - \dfrac{5}{2}{x^2} + 2\end{array}\)
c) \(\begin{array}{l}( - 15{x^6} - 24{x^3}):( - 3{x^2}) = ( - 15{x^6}):( - 3{x^2}) + ( - 24{x^3}):( - 3{x^2})\\ = ( - 15: - 3).({x^6}:{x^2}) + ( - 24: - 3).({x^3}:{x^2})\\ = 5.{x^{6 - 2}} + 8.{x^{3 - 2}} = 5{x^4} + 8x\end{array}\)
2^2 . (-2^3) - ( x + 3 )^3 = 24
4 . (-8) - ( x+ 3 )^3 = 24
=> -32 -( x + 3 )^3 = 24
=> ( x + 3 )^3 = -32 - 24
=> ( x + 3 )^3 = - 56
ĐỀ sai thì phải
a, ĐK: \(x\ne24\)
580 :( x -24 ) =329 -150 : 2
<=> 580 :( x -24 ) = 329 - 75
<=> 580 :( x -24 ) = 254
<=> x - 24 = \(\frac{290}{127}\)
<=> x = \(\frac{3338}{127}\left(TM\right)\)
Vậy \(x=\frac{3338}{127}\)
b, 7 (x-1 ) +35= 25 + 279 :9
<=> 7x - 7 + 35 = 25 + 31
<=> 7x +28 = 56
<=> 7x = 28
<=> x = 4
Vậy x =4
c,4 ( 2x+ 7 ) -3 (3x -2 ) =24
<=> 8x + 28 - 9x + 6 = 24
<=> 34 - x = 24
<=> x = 10
Vậy x = 10
d,( x-1 ) (x-2) =3(x-1)
<=> ( x-1 ) (x-2) - 3(x-1) = 0
<=> (x- 1)(x - 2 - 3) = 0
<=> (x -1)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
Vậy x ={1; 5}
e, (x + 3) + (x + 7) + (x + 11) + ... + (x + 79) = 860
x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
Có tất cả: (79 - 3) : 4 + 1 = 20 số hạng \(\Rightarrow\) có 20x
hay x + 3 + x + 7 + x + 11 + ... + x + 79 = 860
\(\Rightarrow\) 20x + (3 + 7 + ... + 79) = 860
3 + 7 + ... + 79 = (79 + 3) x 20 : 2 = 820
\(\Rightarrow\) 20x + (3 + 7 + ... + 79) = 860
\(\Rightarrow\) 20x + 820 = 860
\(\Rightarrow\) 20x = 40
\(\Rightarrow\) x = 2
Vậy x = 2
Chúc bn học tốt!
1) 4(x+3)2 - 26 = 74
4(x+3)2 = 100
(x+3)2 = 25 = 52 = (-5)2
th1: x + 3 = 5
=> x =2
th2: x + 3 = -5
=> x = -8
Câu 2 tương tự
\(3^{x+2}-3^x=24\)
\(3^x.3^2-3^x.1=24\)
\(3^x.\left(3^2-1\right)=24\)
\(3^x.\left(9-1\right)=24\)
\(3^x.8=24\)
\(3^x=24:8\)
\(3^x=3\)
\(3^x=3^1\)
\(\Rightarrow x=1\)
\(3^{x+2}-3^x=24\)\(\Leftrightarrow3^x.3^2-3^x=24\)
\(\Leftrightarrow3^x\left(3^2-1\right)=24\)\(\Leftrightarrow3^x.8=24\)
\(\Leftrightarrow3^x=3\)\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\(3^{x+2}-3^x=24\)
\(\Rightarrow3^x\left(3^2-1\right)=24\)
\(\Rightarrow3^x.8=24\)
\(\Rightarrow3^x=3\)
\(\Rightarrow x=1\)
Vay x = 1