là walnut nha

walnut

dazzling nha

Blazing

31. Khi cảnh sát đến đó, tên cướp……………………………

• A. đã rời đi
• B. đã rời đi
• C. rời đi
• D. bị bỏ lại

32. Bảy phần mười số người thích giao tiếp mặt đối mặt……khi hẹn hò.

• A. ngôn ngữ
• B. liên hệ/giao tiếp
• C. mã/mật mã
• D. dấu hiệu/biểu tượng

33. Khi cảnh sát đến hiện trường, tài xế ô tô…………………..…

• A. đã rời đi
• B. đã rời đi
• C. rời đi
• D. bị bỏ lại

34. Khi bạn nhìn thấy Jenny, bạn sẽ nhận ra cô ấy. Cô ấy…………….. một chiếc màu đỏ...... (Lưu ý: Câu 34 này bị thiếu từ cuối cùng, ví dụ như "váy", "áo", "mũ", v.v. để hoàn chỉnh ý nghĩa.)-nhờ AI dịch hộ

a\(^2\) + b\(^2\) = (a - b)\(^2\) + 2ab = 1\(^2\) + 2.2 = 1 + 4 = 5

a\(^2\) + b\(^2\) = (a + b)\(^2\) - 2ab = 2\(^2\) + 2.1 = 4 + 2 = 6

A = \(a^2\) + 2\(a^2b\) + 2\(ab^2\) + b\(^2\)

A = (\(a^2+2ab+b^2\)) - 2ab + (2\(a^2b+2ab^2\))

A = (a + b)\(^2\) + 2ab.(a+ b - 1) (1)

Thay a + b = 1 vào biểu thức (1) ta có:

A = 1\(^2\) + 2ab.(1 - 1)

A = 1 + 2.0

A = 1 + 0

A = 1

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

Bài 1:

a: \(\left(\frac{9}{25}-2^2\right):\left(-0,2\right)\)

\(=\left(\frac{9}{25}-4\right):\left(\frac{-1}{5}\right)=\frac{-91}{25}\cdot\frac{-5}{1}=\frac{91}{5}\)

b: \(\left(-\frac15\right)^2+\frac15-2\cdot\left(-\frac12\right)^3-\frac12\)

\(=\frac{1}{25}+\frac15-2\cdot\frac{-1}{8}-\frac12\)

\(=\frac{1}{25}+\frac{5}{25}+\frac14-\frac12=\frac{6}{25}-\frac14=\frac{24}{100}-\frac{25}{100}=-\frac{1}{100}\)

c: \(\left(3-\frac14+\frac23\right)^2:2022^0\)

\(=\left(\frac{36}{12}-\frac{3}{12}+\frac{8}{12}\right)^2=\left(\frac{41}{12}\right)^2=\frac{1681}{144}\)

d: \(2^2\cdot9:\left(3\frac45+0,2\right)\)

\(=4\cdot9:\left(3,8+0,2\right)\)

\(=\frac{36}{4}=9\)

e: \(\left(\frac14+\frac23\right)^2-1\frac13=\left(\frac{3}{12}+\frac{8}{12}\right)^2-\frac43\)

\(=\left(\frac{11}{12}\right)^2-\frac43=\frac{121}{144}-\frac{192}{144}=-\frac{71}{144}\)

f: \(1:\left(-1\frac52+0,5\right)^2\)

\(=1:\left(-\frac72+\frac12\right)^2\)

\(=1:\left(-3\right)^2=\frac19\)

Bài 2:

a: \(-\frac{5}{14}+\frac38-\frac{2}{14}-\frac38+\frac12\)

\(=\left(-\frac{5}{14}-\frac{2}{14}+\frac12\right)+\left(\frac38-\frac38\right)\)

\(=\left(-\frac{7}{14}+\frac{7}{14}\right)+0=0+0=0\)

b: \(\frac{7}{15}-\frac57+\frac{23}{15}+\frac57-\frac35\)

\(=\left(\frac{7}{15}+\frac{23}{15}\right)-\frac35+\left(\frac57-\frac57\right)\)

\(=\frac{30}{15}-\frac35=2-\frac35=\frac75\)

c: \(-\frac25\cdot\frac57+\frac{-2}{5}\cdot\frac97\)

\(=-\frac25\left(\frac57+\frac97\right)=-\frac25\cdot2=-\frac45\)

d: \(\frac{55}{27}+\frac{-21}{5}+\frac{-55}{27}-\frac{-21}{5}\)

\(=\left(\frac{55}{27}-\frac{55}{27}\right)+\left(-\frac{21}{5}+\frac{21}{5}\right)\)

=0+0=0

e: \(\frac57:\left(\frac{15}{8}-\frac14\right)-\frac57:\left(\frac14+\frac12\right)\)

\(=\frac57:\left(\frac{15}{8}-\frac28\right)-\frac57:\left(\frac14+\frac24\right)\)

\(=\frac57:\frac{13}{8}-\frac57:\frac34\)

\(=\frac57\cdot\frac{8}{13}-\frac57\cdot\frac43=\frac57\left(\frac{8}{13}-\frac43\right)=\frac57\cdot\left(\frac{24}{39}-\frac{52}{39}\right)\)

\(=\frac57\cdot\frac{-28}{39}=\frac{5\cdot\left(-4\right)}{39}=-\frac{20}{39}\)

f: \(16\frac27:\left(-\frac35\right)-28\frac27:\left(-\frac35\right)\)

\(=\left(16+\frac27\right)\cdot\frac{-5}{3}-\left(28+\frac27\right)\cdot\frac{-5}{3}\)

\(=-\frac53\left(16+\frac27-28-\frac27\right)=-\frac53\cdot\left(-12\right)=20\)

\(n^2=4\)

\(\Rightarrow\left[\begin{array}{l}n=2\\ n=-2\end{array}\right.\)

vậy n=2 hoặc n=-2

đáp là 2 nhé bạn vì 2 mũ 2 = 4