\(\dfrac{2}{3}-\left|x-2,4\right|=\dfrac{1}{2}\)

\(\left|x-2,4\right|=\dfrac{2}{3}-\dfrac{1}{2}\)

\(\left|x-2,4\right|=\dfrac{1}{6}\)

*) Với \(x\ge2,4\) ta có:

\(x-2,4=\dfrac{1}{6}\)

\(x=\dfrac{1}{6}+2,4\)

\(x=\dfrac{77}{30}\) (nhận)

*) Với \(x< 2,4\) ta có:

\(x-2,4=-\dfrac{1}{6}\)

\(x=-\dfrac{1}{6}+2,4\)

\(x=\dfrac{67}{30}\) (nhận)

Vậy \(x=\dfrac{67}{30};x=\dfrac{77}{30}\)

\(\sqrt{25}.\left(0,4-1\dfrac{1}{2}\right):\left[\left(-2\right)^3:\dfrac{8}{11}\right]\)

\(=5.\left(\dfrac{2}{5}-\dfrac{3}{2}\right):\left(-8:\dfrac{8}{11}\right)\)

\(=5.\left(-\dfrac{11}{10}\right):\left(-11\right)\)

\(=\dfrac{-11}{2}:\left(-11\right)\)

\(=\dfrac{1}{2}\)

\(ƯCLN\left(1;2;3;...;100\right)+ƯCLN\left(1^2;2^2;3^2;...;100^2\right)+ƯCLN\left(1^3;2^3;3^3;...;100^2\right)+...+ƯCLN\left(1^{100};2^{100};3^{100};...;100^{100}\right)\)

\(=1+1+1+...+1\) (100 chữ số 1)

\(=100\)

Lời giải:
$B=1+(5+5^2+5^3)+(5^4+5^5+5^6)+....+(5^{88}+5^{89}+5^{90})$

$=1+5(1+5+5^2)+5^4(1+5+5^2)+....+5^{88}(1+5+5^2)$

$=1+(1+5+5^2)(5+5^4+....+5^{88})$

$=1+31(5+5^4+...+5^{88})\not\vdots 31$
Ta có đpcm.

12,3:7=1,7571428571,94,6:27=3,5037037037

Hữu Ngĩa

H

hiệu khi không giảm số trừ đi 9 là:

     55-9=46

hiệu ban đầu là:

46-14=32

Đ/s:32

50

Lời giải:
ĐKXĐ: $x\geq \frac{-1}{3}$

PT \(\Leftrightarrow 3(\sqrt{3x^2+1}-2)+2(\sqrt{3x+1}-2)+2(x-1)=0\)

\(\Leftrightarrow 3.\frac{3(x^2-1)}{\sqrt{3x^2+1}+2}+2.\frac{3(x-1)}{\sqrt{3x+1}+2}+2(x-1)=0\\ \Leftrightarrow (x-1)\left[\frac{3(x+1)}{\sqrt{3x^2+1}+2}+\frac{2}{\sqrt{3x+1}+2}+2\right]=0\)

Dễ thấy với $x\geq \frac{-1}{3}$ thì biểu thức trong ngoặc vuông luôn dương.

$\Rightarrow x-1=0$

$\Leftrightarrow x=1$ (tm)

(56-27)-(11+28-16)

=29-(39-16)

=29-23

=6

Lời igair:
$(56-27)-(11+27-16)=56-27-11-28+16$

$=(56+16)-(27+11+28)=72-66=6$