Bài 1:tìm x A,”`x^11″`=x B,”`(2×x+1)^3″`=125 C,”`(x-3)^8″`=”`(x-3)^6″` Bài 2: tính tổng Cho A = 3+”`3^2″`+”`3^3″`+…+”`3^2020″` Bài 3 tìm stn 2×a+3=

Đáp án:

1/ a/ $\text{$x=0$ hoặc $x=1$}$

b/ $x=2$

c/ $\text{$x=2$ hoặc $x=3$ hoặc $x=4$}$

2/ $A=\dfrac{3^{2021}-3}{2}$

3/ $a=39$

4/ $x=7$

Giải thích các bước giải:

1/ a/ $x^{11}=x$

$⇒ x^{11}-x=0$

$⇒ x(x^{10}-1)=0$

$⇒ \left[ \begin{array}{l}x=0\\x^{10}=1\end{array} \right.$

$⇒ \left[ \begin{array}{l}x=0\\x=1\end{array} \right.$

b/ $(2x+1)^3=125$

$⇒ 2x+1=5$

$⇒ 2x=4$

$⇒ x=2$

c/ $(x-3)^8=(x-3)^6$

$⇒ (x-3)^8-(x-3)^6=0$

$⇒ (x-3)^6.[(x-3)^2-1]=0$

$⇒ (x-3)^6.(x^2-6x+8)=0$

$⇒ (x-3)^6.(x^2-2x-4x+8)=0$

$⇒ (x-3)^6.(x-2)(x-4)=0$

$⇒ \left[ \begin{array}{l}x=3\\x=2\\x=4\end{array} \right.$

2/ $A=3+3^2+3^3+…..+3^{2020}$

$⇒ 3A=3^2+3^3+3^4+…..+3^{2021}$

$⇒ 3A-A=(3^2+3^3+3^4+…..+3^{2021})-(3+3^2+3^3+…..+3^{2020})$

$⇒ 2A=3^{2021}-3$

$⇒ A=\dfrac{3^{2021}-3}{2}$

3/ $2a+3=3^4$

$⇒ 2a=81-3$

$⇒ 2a=78$

$⇒ a=39$

4/ $(x+1)+(x+2)+(x+3)+…..+(x+100)=5750$

$⇒ 100x+(1+2+3+….+100)=5750$

$⇒ 100x+(100+1).50=5750$

$⇒ 100x=5750-5050$

$⇒ 100x=700$

$⇒ x=7$

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