a)

 \(\begin{array}{l}\left( {13x{\rm{ }}-{\rm{ }}{{12}^2}} \right):{\rm{ }}5{\rm{ }} = {\rm{ }}5\\13x{\rm{ }}-{\rm{ }}{12^2} = 5.5\\13x{\rm{ }}-{\rm{ }}144 = 25\\13x = 25 + 144\\13x = 169\\x = 13\end{array}\)

Vậy \(x = 13\)

b)

\(\begin{array}{l}3x\left[ {{8^2} - 2.\left( {{2^5} - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.\left( {32 - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.31} \right]{\rm{ }} = {\rm{ }}2022\\3x\left( {64 - 62} \right){\rm{ }} = {\rm{ }}2022\\3x.2 = 2022\\6x = 2022\\x = 337\end{array}\)

Vậy \(x = 337.\)

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)

\(a,81\cdot2022+25\cdot2022-6\cdot2022=2022\cdot\left(81+25-6\right)=2022\cdot100=202200\)

\(b,\left(x-1\right)\cdot\frac{2}{3}-\frac{1}{5}=\frac{2}{5}\)

\(\left(x-1\right)\cdot\frac{2}{3}=\frac{3}{5}\)

\(x-1=\frac{9}{10}\)

\(x=\frac{19}{10}\)

Vậy \(x=\frac{19}{10}\)

( Nếu phần b là hỗn số thì mình làm thế kia , còn nếu là nhân thì bạn tham khảo Câu hỏi của lương bảo ngọc - Toán lớp 5 - Học trực tuyến OLM nhé )

81 x 2022 + 25 x 2022 - 6 x 2022

= ( 81 + 25 - 6 ) x 2022

= 100 x 2022

= 202 200

b) \(\left(\text{x - 1}\right)\frac{\text{2}}{\text{3}}-\frac{\text{1}}{\text{5}}=\frac{\text{2}}{\text{5}}\)

\(\frac{\text{3 x }\text{( x - 1 ) }+\text{2}}{\text{3}}=\frac{\text{1}}{\text{5}}+\frac{\text{2}}{\text{5}}=\frac{\text{3}}{\text{5}}\)

=> \(\text{3 x ( x - 1 ) }+\text{2}=\frac{\text{3}}{\text{5}}\text{ x 3 = }\frac{\text{9}}{\text{5}}\)

=> \(\text{3 x ( x - 1 ) }=\frac{\text{9}}{\text{5}}-\text{2}=\frac{\text{-1}}{\text{5}}\)

=> \(\text{ x-1}=\frac{\text{-1}}{\text{5}}:3=\frac{\text{-1}}{\text{15}}\)

=> \(\text{x}=\frac{\text{-1}}{\text{15}}+\text{1 = }\frac{\text{14}}{\text{15}}\)

dạ vâng a off

a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)

b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)

c) \(\dfrac{135}{175}=\dfrac{27}{35}\)

\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)

\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)

\(a,50\%+\dfrac{7}{12}-\dfrac{1}{2}\\ =\dfrac{1}{2}+\dfrac{7}{12}-\dfrac{1}{2}\\ =\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\dfrac{7}{12}\\ =\dfrac{7}{12}\\ b,2022\times67+2022\times43-2022\times10\\ =2022\times\left(67+43-10\right)\\ =2022\times100\\ =202200.\\ c,125-25:3\times12\)

\(=25\times5-25:3\times12\\ =25\times\left(5-\dfrac{1}{3}\right)\times12\\ =25\times\dfrac{14}{3}\times12\\ =1400\)

a,50%+127​−21​=21​+127​−21​=(21​−21​)+127​=127​b,2022×67+2022×43−2022×10=2022×(67+43−10)=2022×100=202200.c,125−25:3×12

=25×5−25:3×12=25×(5−13)×12=25×143×12=1400=25×5−25:3×12=25×(5−31​)×12=25×314​×12=1400

a) 7/12

b) 202200

c) 1400

d) mình ko bt

\(500-\left\{25:\left[15+2.\left(10-20\right)\right]+2030^0\right\}\)

\(=500-\left\{25:\left[15+\left(-20\right)\right]+1\right\}\)

\(=500-\left\{25:\left(-5\right)+1\right\}\)

\(=500-\left(-4\right)\)

\(=504\)

\(2022-\left(-5+2022\right)\)

\(=2022+55-2022\)

\(=\left(2022-2022\right)+55=55\)

vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)

mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)

=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

A

nhầm D

D