giúp gì?

giúp j

giúp cái gì chứ

a) \(\left(x-3\right)^{2022}=\left(x-3\right)^{2023}\)

\(\left(x-3\right)^{2022}\left(x-3-1\right)=0\)

\(\left(x-3\right)^{2022}\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) \(\left(x+1\right)^{2024}+\left(x+1\right)^{2023}=2\)

\(\left[\left(x+1\right)^{2022}-1\right]\left[\left(x+1\right)^{2022}+2\right]=0\)

\(\Leftrightarrow x+1=1\)

\(\Leftrightarrow x=0\)

c) \(\left(x+3\right)^2+\left(x+3\right)=12\)

\(\left[\left(x+3\right)-3\right]\left[\left(x+3\right)+4\right]=0\)

\(x\left(x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\end{matrix}\right.\)

1B

2C

3A

4D

5C

6A

Chào em, em tham khảo nhé:

1. B

2. C

3. A

4. D

5. C

6. A

Chúc em học tốt.

1B

2C

3A

4D

5C

6A

còn cần kh á

a) \(k=-5\)

b) 

Tìm x xong rồi tìm y

3 thì làm kiểu gì cũng được

Bài 3:

Hình 1:

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(x+60^0+40^0=180^0\)

=>\(x=180^0-100^0=80^0\)

Hình 2:

Xét ΔABD có \(\widehat{ADC}\) là góc ngoài tại đỉnh D

nên \(\widehat{ADC}=\widehat{ABD}+\widehat{BAD}=110^0\)

=>\(y=110^0\)

ΔDAC cân tại D

=>\(\widehat{DAC}=\widehat{DCA}=\dfrac{180^0-\widehat{ADC}}{2}\)

=>\(x=\dfrac{180^0-110^0}{2}=35^0\)

Hình 3:

Ta có: \(\widehat{CAB}+60^0=180^0\)

=>\(\widehat{CAB}=180^0-60^0=120^0\)

Xét ΔCAB có \(\widehat{CAB}+\widehat{ABC}+\widehat{ACB}=180^0\)

=>\(x+2x+120^0=180^0\)

=>\(3x=60^0\)

=>\(x=20^0\)

Hình 4:

Ta có: \(\widehat{ADC}+\widehat{ADB}=180^0\)

=>\(\widehat{ADB}=180^0-80^0=100^0\)

Xét ΔADB có

\(\widehat{ADB}+\widehat{ABD}+\widehat{BAD}=180^0\)

=>\(x+60^0+100^0=180^0\)

=>\(x=20^0\)

ta có: \(\widehat{ACD}+135^0=180^0\)

=>\(\widehat{ACD}=180^0-135^0=45^0\)

Xét ΔACD có \(\widehat{ACD}+\widehat{ADC}+\widehat{DAC}=180^0\)

=>\(y+80^0+45^0=180^0\)

=>\(y=55^0\)

1 had it fixed

2 have it replaced

3 had a new monitor deliver

4 had it tested

5 have the software installed

6 had all those problems solved

Bài 5

1 be - the past participle

2 the past participle

3 of

4 recycled

5 fall

6 environment

a: \(18\cdot7+65:13\)

=126+5

=131

b: \(785-\dfrac{323+148}{3}+2781\)

\(=785+2781-\dfrac{471}{3}\)

=3566-157

=3409

c: \(703-\dfrac{140}{42+28}-17^6\cdot17^9:17^{13}\)

\(=703-\dfrac{140}{70}-17^2\)

\(=703-289-2\)

=703-291

=412

e: \(\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(=\dfrac{9^3\left(2^3\cdot9+45\right)}{9^2\cdot\left(10-1\right)}=\dfrac{9^3}{9^3}\cdot\left(8\cdot9+45\right)\)

=72+45

=117