a) Với \(2x+1\ge0\Rightarrow x\ge\frac{-1}{2}\) thì \(2x+1-2=1\Rightarrow2x+1=3\Rightarrow2x=2\Rightarrow x=1\) (nhận)

    Với 2x + 1  <0 => x < -1/2 thì \(-\left(2x+1\right)-2=1\Rightarrow-2x-1-2=1\Rightarrow-2x=4\Rightarrow x=-2\) (nhận)

Vậy x = {-2;1}

b) Với \(3x-1\ge0\Rightarrow x\ge\frac{1}{3}\) thì \(3x-1=1-3x\Rightarrow6x=2\Rightarrow x=\frac{1}{3}\) (nhận)

    Với 3x - 1 < 0 => x < 1/3 thì \(-\left(3x-1\right)=1-3x\Rightarrow-3x+1=1-3x\Rightarrow0x=0\)  (nhận)

    Vậy x = 1/3 hoặc x vô số nghiệm với x < 1/3

a) | 2x + 1| - 2 = 1

| 2x + 1| = 3

TH1: 2x + 1 = 3

2x = 2

x = 1

TH2: 2x + 1 = -3

2x = -4

x = -2

KL: x = 1 hoặc x = - 2

b) \(\frac{-5\cdot7^5+7^4}{7^6\cdot10-2\cdot7^5}\)

\(=\frac{-35\cdot7^4+7^4}{7^5\cdot70-2\cdot7^5}\)

\(=\frac{7^4\left(-35+1\right)}{7^5\left(70-2\right)}\)

\(=\frac{7^4\cdot\left(-34\right)}{7^5\cdot68}\)

\(=\frac{-1}{14}\)

Chắc sai =))

Lời giải:
a.

$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$

b.

$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$

c.

$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$

$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$

d.

$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$

$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$

e.

$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$

$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$

d)

\(\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}\\ =\dfrac{3^{29}.2^6.2^2}{3^{24}.3^5.2^6}\\ =\dfrac{3^{29}.2^6.4}{3^{29}.2^6}\\ =4\)

e)

\(\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}\\ =\dfrac{2^{21}.5^3.3^4}{2^3.2^{18}3^4.5}\\ =\dfrac{2^{21}.5.5^2.3^4}{2^{21}.3^4.5}\\ =5^2\\ =25\)

f)

\(=\dfrac{24\left(315+561+124\right)}{\dfrac{\left(1+99\right).50}{2}-500}\\ =\dfrac{24.1000}{2500-500}\\ =12\)

\(a,\dfrac{-14.15}{21.\left(-10\right)}=\dfrac{-7.2.3.5}{7.3.\left(-2\right).5}=1\)

\(b,\dfrac{5.7-7.9}{7.2+6.7}=\dfrac{7\left(5-9\right)}{7\left(2+6\right)}=\dfrac{-4}{8}=-\dfrac{1}{2}\)

\(c,\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}=\dfrac{-7.\left(3+4\right)}{7\left(-5-2\right)}\)

\(=\dfrac{\left(-7\right).7}{7.\left(-7\right)}=1\)

\(d,\dfrac{3^9.3^{20}.2^8}{3^{24}.243.2^6}=\dfrac{3^{29}.2^8}{3^{24}.3^5.2^6}=\dfrac{3^{29}.2^8}{3^{29}.2^6}=2^2=4\)

\(e,\dfrac{2^{15}.5^3.2^6.3^4}{8.2^{18}.81.5}=\dfrac{2^{21}.3^4.5^3}{2^{18}.2^3.3^4.5}=\dfrac{2^{21}.3^4.5^3}{2^{21}.3^4.5}=5^2=25\)

\(f,\dfrac{24.315+3.561.8+4.124.6}{1+3+5+...+97+99-500}\)

\(=\dfrac{24.315+24.561+24.124}{1+3+5+...+97+99-500}\)

\(=\dfrac{24\left(315+561+124\right)}{1+3+5+...+97+99-500}\)

\(=\dfrac{24.1000}{1+3+5+...+97+99-500}\) (1)

Đặt A = 1 + 3 + 5 + ... + 97 + 99

Số số hạng trong A là: (99 - 1) : 2 + 1 = 50 (số)

Tổng A bằng: (99 + 1) . 50 : 2 = 2500

Thay A = 2500 vào biểu thức (1), ta được:

\(\dfrac{24.1000}{2500-500}=\dfrac{24.1000}{2.1000}=12\)

a)

\(\dfrac{\left(-14\right).15}{21.\left(-10\right)}\\ =\dfrac{-7.2.3.5}{7.3.-2.5}\\=\dfrac{7.2.3.5}{7.2.3.5}\\ =1\)

b)

\(\dfrac{5.7-7.9}{7.2+6.7}\\ =\dfrac{7\left(5-9\right)}{7\left(2+6\right)}\\ =\dfrac{-4}{8}\\ =\dfrac{-2.2}{2.4}\\ =-\dfrac{1}{2}\)

c)

\(\dfrac{\left(-7\right).3+2.\left(-14\right)}{\left(-5\right).7-2.7}\\ =\dfrac{-7.3+2.-7.2}{7\left(-5-2\right)}\\ =\dfrac{-7\left(3+4\right)}{7.-7}\\ =\dfrac{7}{7}\\ =1\)

trình bày đày đủ giúp em ạ

\(a,\dfrac{5}{3}\times\dfrac{4}{5}=\dfrac{4}{3}\) (phần rút gọn : bỏ 5)

\(b,\dfrac{2}{3}\times\dfrac{3}{7}=\dfrac{2}{7}\) (phần rút gọn : bỏ 3)

\(c,\dfrac{7}{13}\times\dfrac{13}{7}=1\) (phần rút gọn : bỏ 7 và 13)

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

\(=\sqrt[3]{1+3\sqrt{2}+3\sqrt{2^2}+2\sqrt{2}}-\sqrt[3]{2\sqrt{2}-3\sqrt{2^2}+3\sqrt{2}-1}\)

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

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

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

\(=1+\sqrt{5}-\left(\sqrt{5}-1\right)\)

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

Vậy...