phân tích đa thức thành nhân tử dạng đặt biến phụ
1, (x-1)(x-3)(x-5)(x-7)-20
2, (x-1)((x+2)(x+3)(x+6)-28
3, x(x-1)(x+1)(x+2)-3
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(x+1)(x+2)(x+3)(x+4)-24
= (x^2 + 5x + 4)(x^2 + 5x + 6) - 24
đặt x^2 + 5x + 5 = a
ta có : (a - 1)(a + 1) - 24 = a^2 - 1 - 24
= a^2 - 25
= (a - 5)(a+5)
= (x^2 + 5x + 5 - 5)(x^2 + 5x + 5 + 5)
= (x^2 + 5x)(x^2 + 5x + 10)
= x(x + 5)(x^2 + 5x + 10)
x(x+1)(x+2)(x+3)+1
= (x^2 + 3x)(x^2 + 3x + 2) + 1
đặt x^2 + 3x + 1 = a
ta có : (a - 1)(a+1) + 1 = a^2 - 1 + 1 = a^2
= (x^2 + 3x + 1)^2
(x+1)(x+2)(x+3)(x+4)−24f,(x+1)(x+2)(x+3)(x+4)−24
=(x+1)(x+4)(x+2)(x+3)−24=(x+1)(x+4)(x+2)(x+3)−24
=(x2+5x+4)(x2+5x+6)−24=(x2+5x+4)(x2+5x+6)−24
Đặt t=x2+5x+4t=x2+5x+4 , ta có
t(t+2)−24t(t+2)−24
=t2+2t−24=t2+2t−24
=(t2+2t+1)−25=(t2+2t+1)−25
=(t+1)2−52=(t+1)2−52
=(t+1−5)(t+1+5)=(t+1−5)(t+1+5)
=(t−4)(t+6)=(t−4)(t+6)
=(x2+5x+4−4)(x2+5x+4+6)=(x2+5x+4−4)(x2+5x+4+6)
=(x2+5x)(x2+5x+10)
x(x+1)(x+2)(x+3)+1x(x+1)(x+2)(x+3)+1
=[x(x+3)][(x+1)(x+2)]+1=[x(x+3)][(x+1)(x+2)]+1
=(x2+3x)(x2+2x+x+2)+1=(x2+3x)(x2+2x+x+2)+1
=(x2+3x)(x2+3x+2)+1=(x2+3x)(x2+3x+2)+1(1)
Đặt x2+3x=t⇒x2+3x+2=t+2x2+3x=t⇒x2+3x+2=t+2
Do đó (1)=t(t+2)+1=t2+2t+1=(t+1)2(1)=t(t+2)+1=t2+2t+1=(t+1)2(*)
Vì t=x2+3xt=x2+3x nên
(*)=(x2+3x+1)2
Bài 1.
Tổng số Electron là 50
\(\rightarrow2Z_x+3Z_y=50\left(1\right)\)
Mặt khác hiệu số Proton là 5
\(\rightarrow Z_x-Z_y=5\left(2\right)\)
Từ (1) và (2) \(\rightarrow Z_x=13\) và \(Z_y=8\)
Vậy Y là Oxi và X là nhôm
Bài 2.
\(M_X=\frac{46,8}{0,45}=104g/mol\)
Đặt CTHH của \(X=Mg_xS_yO_z\)
\(\rightarrow m_{Mg}:m_S:m_O=\frac{3x}{24}:\frac{4y}{32}:\frac{6z}{16}\)
\(\rightarrow m_{Mg}:m_S:m_O=x:y:3z\)
Chọn tỉ lệ tối giản \(\rightarrow\hept{\begin{cases}x=y=1\\z=3\end{cases}}\)
Vậy CTHH là \(MgSO_3\)
Bài 1: Tổng số Electron là 50
\(\rightarrow2Z_x+3Z_y=5\left(1\right)\)
Mặt khác hiệu số Proton là 5
\(\rightarrow Z_x-X_y=5\left(2\right)\)
\(\Rightarrow Z_x=13;Z_y=8\)
\(\Rightarrow\)Vây Y là oxy còn X là nhôm
link nè vào đi https://olm.vn/bai-viet/nguyen-gia-han-co-be-hieu-thao%F0%9F%92%96%F0%9F%92%96-163860
1. He often drives vety _____________.(dangerous / dangerously)
2. The music they played music ___________.(soft / softly)
3.The teacher looked very _____________ this morning (angry / angrily)
4. This is a very ______ pen. (good / well)
5. The song she sang at the party was very _____. (bad / badly)
* Nếu sai thì thông cảm ạ :) *
1. He often drives vety _____________.(dangerous / dangerously)
2. The music they played music ___________.(soft / softly)
3.The teacher looked very _____________ this morning (angry / angrily)
4. This is a very ______ pen. (good / well)
5. The song she sang at the party was very _____. (bad / badly)
1.Lam said Hoa should do exercise regularly
2.Lam's mother told her to wash her clothes.
3.He used to drive carelessly when he was small.
4.Test in class 8D is easier than test in class 8E.
1 , .Lam said Hoa should do exercise regularly
2 , Lam's mother told her to wash her clothes.
3 , He used to drive carelessly when he was small
4 , Test in class 8D is easier than test in class 8E
Rewrite the sentences:
Despite in spite of being a little overweight, he is actually quite fit.
=> Even though he is a little overweight, he is actually quite fit
/ Câu này giúp được Ri nè:3 /
Hc tốt
1. (x-1)(x-3)(x-5)(x-7)-20=0
<=> (x-1)(x-7)(x-3)(x-5)-20=0
<=> (x^2-8x+7)(x^2-8x+15)-20=0
Đặt x^2-8x+7=a => x^2-8x+15= a+8
=> a(a+8)-20=0
<=> a^2+8a-20=0
<=>(a^2+8a+16)-36=0
<=> (a+4)^2=36
=> {a+4=6a+4=−6{a+4=6a+4=−6
<=>{a=2a=−10{a=2a=−10
*a=2 => x^2-8x+7=2
<=> x^2-8x+5=0
<=>(x^2-8x+16)-11=0
<=>(x-4)^2=11
<=>x-4=√11
<=> x=√11 +4
*a=-10 => x^2-8x+7=-10
<=> x^2-8x+17=0
<=> (x^2-8x+16)+1=0
<=> (x-4)^2=-1 (PT vô nghiệm)
Vậy pt có nghiệm x=√11 +4
mk chỉ biết vậy thôi
3, \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)-3=\left(x^2+x\right)\left(x^2+x-2\right)-3\)
Đặt \(x^2+x=t\)
\(t\left(t-2\right)-3=t^2-2t-3=\left(t-3\right)\left(t+1\right)\)
Theo cách đặt \(\left(x^2+x-3\right)\left(x^2+x+1\right)\)