1: =>(2x-1)^3=0

=>2x-1=0

=>x=1/2

2: =>(x-2)^3=27

=>x-2=3

=>x=5

3: =>(x-4)^2=5(4-x)^3

=>(x-4)^2=-5(x-4)^3

=>(x-4)^2+5(x-4)^3=0

=>(x-4)^2*[5(x-4)+1]=0

=>(x-4)^2(5x-19)=0

=>x=4 hoặc x=19/5

4: =>(2-x)^3+6x(2-x)=0

=>(2-x)[(2-x)^2+6x]=0

=>(2-x)(x^2-4x+4+6x]=0

=>(2-x)(x^2+2x+4)=0

=>2-x=0

=>x=2

5: =>x^3+3x^2+3x+1-x^3+3x^2-3x+1-6(x^2-2x+1)=-10

=>6x^2+2-6x^2+12x-6=-10

=>12x-4=-10

=>12x=-6

=>x=-1/2

6: =>27-27x+9x^2-x^3-x^3-9x^2-27x-27=36x^2-54x

=>36x^2-54x=-2x^3-54x

=>-2x^3=36x^2

=>-2x^3-36x^2=0

=>-2x^2(x+18)=0

=>x=0 hoặc x=-18

17:

HB/HC=1/4

=>HB/1=HC/4=k

=>HB=k; HC=4k
Xét ΔABC vuông tại A có AH là đường cao

nên AH^2=HB*HC

=>4k^2=14^2=196

=>k^2=49

=>k=7

=>HB=7cm; HC=28cm

BC=7+28=35cm

18:

AH=14cm; HB/HC=1/4

=>HB=7cm; HC=28cm

BC=7+28=35cm

AC=căn HC*BC=14căn 5(cm)

AB=căn BH*BC=7căn 5(cm)

C=7căn 5+14căn 5+35=21căn5+35(cm)

a) Ta có :

\(27^{27}>27^{26}=\left(27^2\right)^{13}=729^{13}>243^{13}\)

\(\Rightarrow27^{27}>243^{13}\)

\(\Rightarrow-27^{27}< -243^{13}\)

\(\Rightarrow\left(-27\right)^{27}< \left(-243\right)^{13}\)

b) \(\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{8}\right)^{26}=\left(\dfrac{1}{8^2}\right)^{13}=\left(\dfrac{1}{64}\right)^{13}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(\dfrac{1}{8}\right)^{25}>\left(\dfrac{1}{128}\right)^{13}\)

\(\Rightarrow\left(-\dfrac{1}{8}\right)^{25}< \left(-\dfrac{1}{128}\right)^{13}\)

c) \(4^{50}=\left(4^5\right)^{10}=1024^{10}\)

\(8^{30}=\left(8^3\right)^{10}=512^{10}< 1024^{10}\)

\(\Rightarrow4^{50}>8^{30}\)

d) \(\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{9}\right)^{12}< \left(\dfrac{1}{27}\right)^{12}\)

\(\Rightarrow\left(\dfrac{1}{9}\right)^{17}< \left(\dfrac{1}{27}\right)^{12}\)

a) Ta có :

$27^{27}>27^{26}={\left(27^2\right)}^{13}=729^{13}>243^{13}$

$\Rightarrow 27^{27}>243^{13}$

$\Rightarrow -27^{27}<-243^{13}$

$\Rightarrow {\left(-27\right)}^{27}<{\left(-243\right)}^{13}$

b) ${\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{26}={\left(\genfrac{}{}{0ex}{}{1}{8^2}\right)}^{13}={\left(\genfrac{}{}{0ex}{}{1}{64}\right)}^{13}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}>{\left(\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

$\Rightarrow {\left(-\genfrac{}{}{0ex}{}{1}{8}\right)}^{25}<{\left(-\genfrac{}{}{0ex}{}{1}{128}\right)}^{13}$

c) $4^{50}={\left(4^5\right)}^{10}=1024^{10}$

$8^{30}={\left(8^3\right)}^{10}=512^{10}<1024^{10}$

$\Rightarrow 4^{50}>8^{30}$

d) ${\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{12}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

$\Rightarrow {\left(\genfrac{}{}{0ex}{}{1}{9}\right)}^{17}<{\left(\genfrac{}{}{0ex}{}{1}{27}\right)}^{12}$

\(x-\dfrac{3}{5}=\dfrac{2}{7}+\dfrac{1}{6}\)

\(\Leftrightarrow x-\dfrac{3}{5}=\dfrac{19}{42}\)

\(\Leftrightarrow x=\dfrac{19}{42}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{221}{210}\)

=>x-3/5=12/42+7/42=19/42

=>x=19/42+3/5=95/210+126/210=221/210

a) \(A=10^{100}+5\)

- Tận cùng A là số 5 \(\Rightarrow A⋮5\)

- Tổng các chữ số của A là \(1+5=6⋮3\Rightarrow A⋮3\) \(\)

\(\Rightarrow dpcm\)

b) \(B=10^{50}+44\)

- Tận cùng B là số 4 là số chẵn \(\Rightarrow B⋮2\)

- Tổng các chữ số của B là \(1+4+4=9⋮9\Rightarrow B⋮9\)

\(\Rightarrow dpcm\)

Lời giải:

a. Gọi $d$ là ƯCLN $(n+3, 2n+7)$

$\Rightarrow n+3\vdots d$ và $2n+7\vdots d$

$\Rightarrow 2n+7-2(n+3)\vdots d$

Hay $1\vdots d$

$\Rightarrow d=1$

Vậy $n+3, 2n+7$ nguyên tố cùng nhau, nên $\frac{n+3}{2n+7}$ tối giản.

b.

Gọi $d$ là ƯCLN $(4n+6, 6n+7)$

$\Rightarrow 4n+6\vdots d; 6n+7\vdots d$

$\Rightarrow 3(4n+6)-2(6n+7)\vdots d$
$\Rightarrow 4\vdots d$

Mặt khác, vì $6n+7\vdots d$ mà $6n+7$ lẻ nên $d$ lẻ.

$\Rightarrow d=1$

$\Rightarrow \frac{4n+6}{6n+7}$ tối giản.

Lời giải:
Tỉ số giữa tử và mẫu: $\frac{12}{18}=\frac{2}{3}$

Mẫu số là 35 thì tử số là: $35.\frac{2}{3}=\frac{70}{3}$ (không phải số tự nhiên) - nghe không hợp lý lắm. Bạn xem lại đề.

\(a-b=\sqrt{2}+1;b-c=\sqrt{2}-1\Rightarrow a-c=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)

Ta có: \(A=a^2+b^2+c^2-ab-bc-ca=\dfrac{2a^2+2b^2+2c^2-2ab-2bc-2ca}{2}\)

\(=\dfrac{(a^2-2ab+b^2)+\left(b^2-2bc+c^2\right)+\left(c^2-2ca+a^2\right)}{2}\)

\(=\dfrac{\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2}{2}\)

\(=\dfrac{\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2+\left(2\sqrt{2}\right)^2}{2}\)

\(=\dfrac{3+2\sqrt{2}+3-2\sqrt{2}+8}{2}=\dfrac{14}{2}=7\)

Vậy, A=7