CÁC BẠN GIÚP MÌNH MẤY CÂU NÀY NHÉ!

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Giải lại (lần này giải 1 trường hợp thôi, kẻo lại bị troll ức chế:v)

PT (2) \(\Leftrightarrow\left(x+1-\sqrt{y+4}\right)\left(x+\sqrt{y+4}-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=\sqrt{y+4}\left(3\right)\\x+\sqrt{y+4}=3\left(4\right)\end{cases}}\). 

*Xét (3): Thêm điều kiện \(x\ge-1\). (3) \(\Leftrightarrow y=x^2+2x-3\) (bình phương lên:v)

Thay vào PT (1) \(\Leftrightarrow\left(1-x\right)\left(x+2\right)\left(x^4+4x^3-x^2-12x+9\right)=0\)

Vì x + 2 > 0 và \(\left(x^4+4x^3-x^2-12x+9\right)\)

\(=\frac{\left(x+5\right)\left[4\left(x-1\right)^2\left(x+2\right)+1\right]+x^2\left(x+1\right)\left(x^2+2x-2\right)^2}{\left(x+1\right)\left(x^2+1\right)+4}>0\)

Do đó x = 1. Thay vào (3) suy ra y = 0.

(4) giải tương tự cũng cho nghiệm x = 1; y= 0

bác này phải ngủ

Giải xong bấm thì bảo plese sign! Bực:( Giải cả 2 trang giấy dày đặc:(

\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\) 

        2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)

        2 - 3 \(\times\) y = \(\dfrac{8}{15}\)

             3 \(\times\) y = 2 - \(\dfrac{8}{15}\)

             3 \(\times\) y = \(\dfrac{22}{15}\)

                   y  = \(\dfrac{22}{15}\) : 3 

                   y = \(\dfrac{22}{45}\)

1.a.

\(\left(x+3\right)\left(x-2\right)< 0\)

\(TH1:\hept{\begin{cases}x+3< 0\\x-2>0\end{cases}}\Rightarrow\hept{\begin{cases}x< -3\\x>2\end{cases}}\)

\(TH2:\hept{\begin{cases}x+3>0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x>-3\\x< 2\end{cases}}}\)

không biết có đúng không nữa!

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)

mà A\(\le0\)

​​\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\)​ phải bằng 0 đê thỏa mãn điều kiện

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy....

b;c)I hệt câu a nên làm tương tự nhá

d)

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)

B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)

Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)

Vậy....

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|\ge0\\\left|y+\dfrac{4}{3}\right|\ge0\\\left|z+\dfrac{7}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\Rightarrow x=-\dfrac{9}{2}\\\left|y+\dfrac{4}{3}\right|=0\Rightarrow y=-\dfrac{4}{3}\\\left|z+\dfrac{7}{2}\right|=0\Rightarrow z=-\dfrac{7}{2}\end{matrix}\right.\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\\left|y-\dfrac{2}{5}\right|\ge0\\\left|z+\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}\right|=0\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}\right|=0\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|\ge0\\ \left|y+\dfrac{1980}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\Rightarrow x=-\dfrac{19}{5}\\ \left|y+\dfrac{1980}{1975}\right|=0\Rightarrow y=-\dfrac{1980}{1975}\\\left|z-2004\right|=0\Rightarrow z=2004\end{matrix}\right.\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\ \left|y-\dfrac{1}{5}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{1}{5}\right|=0\Rightarrow y=\dfrac{1}{5}\\\left|x+y+z\right|=0\Rightarrow z+-\dfrac{11}{20}=0\Rightarrow z=\dfrac{11}{20}\end{matrix}\right.\)

Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\Rightarrow A\ge0\)

Mà ĐK đề là \(A\le0\)

\(\Rightarrow A=0\)

\(\left[{}\begin{matrix}\left|x+\dfrac{3}{4}=0\right|\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}=0\right|\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}=0\right|\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

Các câu còn lại tương tự nhé

a, y \(\times\) \(\dfrac{4}{3}\) = \(\dfrac{16}{9}\)

    y         =    \(\dfrac{16}{9}\) : \(\dfrac{4}{3}\)

    y         = \(\dfrac{4}{3}\)

b, ( y - \(\dfrac{1}{2}\)) + 0,5 = \(\dfrac{3}{4}\)

    y - 0,5 + 0,5 = \(\dfrac{3}{4}\)

   y                   = \(\dfrac{3}{4}\)

c, \(\dfrac{4}{5}-\dfrac{2}{5}y\) = 0,2

   0,8 - 0,4y = 0,2

           0,4y = 0,8 - 0,2

           0,4y  = 0,6

               y = 1,5

d, (y + \(\dfrac{3}{4}\)) \(\times\) \(\dfrac{5}{7}\) = \(\dfrac{10}{9}\)

    y + \(\dfrac{3}{4}\)           = \(\dfrac{10}{9}\) : \(\dfrac{5}{7}\)

   y + \(\dfrac{3}{4}\)            = \(\dfrac{14}{9}\)

y                    = \(\dfrac{14}{9}\) - \(\dfrac{3}{4}\)

 y                   =   \(\dfrac{29}{36}\)

e, y : \(\dfrac{5}{4}\)         = \(\dfrac{9}{5}\)  + \(\dfrac{1}{2}\)

   y : \(\dfrac{5}{4}\)         =   \(\dfrac{23}{10}\)

  y                =      \(\dfrac{23}{10}\)

  y               =   \(\dfrac{23}{8}\)

f, y \(\times\) \(\dfrac{1}{2}\) + \(\dfrac{3}{2}\) \(\times\) y   = \(\dfrac{4}{5}\)

   y \(\times\) ( \(\dfrac{1}{2}+\dfrac{3}{2}\))      =  \(\dfrac{4}{5}\)

   2y                       = \(\dfrac{4}{5}\)

    y                        = \(\dfrac{2}{5}\)

\(\left(y-2\right)\left(y-3\right)+\left(y-2\right)-1=0\)

\(\Leftrightarrow\left(y-2\right)\left(y-3\right)+\left(y-3\right)=0\)

\(\Leftrightarrow\left(y-3\right)^2=0\)

\(\Leftrightarrow y=3\)

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{0;-3;2\right\}\)

Bài làm

Vì ( y - 2 ) . ( y - 3 ) + ( y - 2 ) - 1 = 0

=> ( y - 2 ) = 0 hoặc ( y - 3 ) + ( y - 2 ) - 1 = 0

=> y = 2 hoặc y = 3 

Vậy y = 2 hoặc y = 3

~ Mấy câu còn lại làm tương tự. Làm theo mẫu câu a . b = 0 , => a = 0 hoặc b = 0. ~
# Chúc bạn học tốt # 

\(2\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\)

\(\Leftrightarrow x\in\left\{2;-3\right\}\)

\(\left(x-7\right)\left(x+3\right)=\left(x+3\right)\left(2x-9\right)\)

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)-\left(x+3\right)\left(2x-9\right)=0\)

\(\Leftrightarrow\left(x-7-2x+9\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\)

\(\Leftrightarrow x\in\left\{2;-3\right\}\)