\(\left(x+2\right)^2-\left(2x+1\right)\left(x+2\right)=0\\ < =>\left(x+2\right)\left[\left(x+2\right)-\left(2x+1\right)\right]=0\\ < =>\left(x+2\right)\left(x+2-2x-1\right)=0\\ < =>\left(x+2\right)\left(1-x\right)=0\\ < =>\left[{}\begin{matrix}x+2=0\\1-x=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Vậy: ... 

\(\left(x+2\right)^2-\left(2x+1\right)\left(x+2\right)=0\)

=>(x+2)(x+2-2x-1)=0

=>(x+2)(-x+1)=0

=>\(\left[{}\begin{matrix}x+2=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

\(42+x^3:5=19\)

=>\(x^3:5=19-42=-23\)

=>\(x^3=-115\)

=>\(x=\sqrt[3]{-115}\)

sai rồi

Tôi ghi: x3 có phải là:\(x^3\) đâu

Ta có: \(\widehat{xMy'}=\widehat{x'My}\)(hai góc đối đỉnh)

mà \(\widehat{xMy'}=90^0\)

nên \(\widehat{x'My}=90^0\)

Ta có: \(\widehat{xMy'}+\widehat{x'My'}=180^0\)(hai góc kề bù)

=>\(\widehat{x'My'}=180^0-90^0=90^0\)

Ta có: \(\widehat{xMy}=\widehat{x'My'}\)(hai góc đối đỉnh)

mà \(\widehat{x'My'}=90^0\)

nên \(\widehat{xMy}=90^0\)

Bạn kiểm tra lại đề nhé.

\(\dfrac{2}{3}-\left[-\dfrac{7}{4}-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}+\dfrac{7}{4}+\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\)

\(=\dfrac{8}{12}+\dfrac{21}{12}+\dfrac{6}{12}+\dfrac{3}{8}\)

\(=\dfrac{35}{12}+\dfrac{3}{8}=\dfrac{70}{24}+\dfrac{9}{24}=\dfrac{79}{24}\)

\(\dfrac{2}{3}-\left[\dfrac{-7}{4}-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\\ =\dfrac{2}{3}-\left[\dfrac{-7}{4}-\left(\dfrac{4}{8}+\dfrac{3}{8}\right)\right]\\ =\dfrac{2}{3}-\left(\dfrac{-7}{4}-\dfrac{7}{8}\right)\\ =\dfrac{2}{3}-\left(\dfrac{-14}{8}-\dfrac{7}{8}\right)\\ =\dfrac{2}{3}+\dfrac{21}{8}\\ =\dfrac{16}{24}+\dfrac{63}{24}\\ =\dfrac{79}{24}\)

\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}\)

=\(\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)\)

= \(0,9+\left(\dfrac{2}{6}+\dfrac{1}{6}\right)+\left(\dfrac{25}{35}-\dfrac{4}{35}\right)\)

= \(0,9+\dfrac{3}{6}+\dfrac{21}{35}\)

= `0,9 +0,5 + 0,6`

= `2`

`(-1/27) . 3/7 + 5/9 . (-3/7)`

`1/27 . (-3/7) + 5/9 . (-3/7)`

`(1/27 + 5/9) . (-3/7)`

`16/27 . (-3/7)`

`-16/63`

(\(\dfrac{3}{7}\)+(\(-\dfrac{3}{7}\))). \(\left(-\dfrac{1}{27}\right)\).\(\dfrac{5}{9}\)

= 0.\(\left(-\dfrac{1}{27}\right)\).​\(\dfrac{5}{9}\)

=0

Số kg sắt quặng A :

\(80x\dfrac{40}{100}=32\left(kg\right)\)

Tổng số kg sắt quặng B :

\(32+20=52\left(kg\right)\)

Tổng số kg quặng B gồm cả sắt :

\(80+20=100\left(kg\right)\)

Phần trăm sắt quặng B :

\(\dfrac{52}{100}x100\%=52\%\)

Đáp số : \(52\%\)

Quặng B nặng số kg là: 

`80 + 20= 100 (kg)`

Nung `80`kg quặng A thu được số kg sắt là: 

`80 : 100` x `40 = 32 (kg)`

Số kg sắt có trong quặng B là: 

`32+ 20 = 52 (kg)`

Quặng B chứa % sắt là: 

`52 : 100 `  x `100 = 52`% (sắt)

Đáp số: ....

\(\left(-\dfrac{3}{5}\right)^2.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)

\(=\dfrac{9}{25}.\dfrac{5}{11}+\dfrac{9}{25}.\left(-\dfrac{16}{11}\right)\)

\(=\dfrac{9}{25}.\left[\dfrac{5}{11}+\left(-\dfrac{16}{11}\right)\right]\)

\(=\dfrac{9}{25}.\left(-1\right)\)

\(=-\dfrac{9}{25}\)