tự đi mà làm nha

X=+-5 câu a

​

\(1.x^2+11x=0\)

\(\Leftrightarrow x\left(x+11\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+11=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-11\end{cases}}\)

\(2.\left(x^2-1\right)\left(x^2-9\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+9\right)\left(x-9\right)=0\)

chia thành 4 TH :

\(TH1:X-1=0\)

\(\Leftrightarrow x=1\)

\(TH2:x+1=0\)

\(\Leftrightarrow x=-1\)

\(TH3:X+9=0\)

\(\Leftrightarrow X=-9\)

\(TH4:x-9=0\)

\(\Leftrightarrow x=9\)

Kết luận ....

\(3.\left(\left|x+1\right|-5\right)\left(x^2-9\right)\)

\(\Leftrightarrow\left(\left|x+1\right|-5\right)\left(x-3\right)\left(x+3\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|-5=0\\x-3=0\\x+3=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left|x+1\right|=5\\x=3\\x=-3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+1=+_-5\Leftrightarrow x+1=5,x+1=-5\Leftrightarrow x=4,x=-6\\x=3x\\x=-3\end{cases}}\)

kết luận x=.....

\(4.\left(3x-16\right)⋮\left(x+2\right)\)

\(\Leftrightarrow\left(3x+6\right)-22\)

\(\Leftrightarrow3\left(x+2\right)-22⋮\left(x+2\right)\)

Vì\(\left(x+2\right)⋮\left(x+2\right)\)

\(\Rightarrow\left(3x-16\right)⋮\left(x+2\right)\)

Kết luận x=.....

\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)

=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)

\(\Rightarrow7x< 36< 7x+7\)

\(\Rightarrow x< \dfrac{36}{7}< x+1\)

\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)

\(\Rightarrow x=5\)

tik cho mình nhé

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\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)

\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)

7\(x\) < 36 < 63\(x\) + 7

⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)

\(\dfrac{29}{63}\)<  \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}

⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)

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

=> \(\orbr{\begin{cases}\left(x+1\right)^2=0\\\left(x-2\right)^2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x+1=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)

b) \(\left(x-9\right)^5\left(x-5\right)^8=0\)

=> \(\orbr{\begin{cases}\left(x-9\right)^5=0\\\left(x-5\right)^8=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-9=0\\x-5=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=9\\x=5\end{cases}}\)

câu 1;

bạn nhóm 2 cái đầu với 2 cái cuối  đặt nhân tử chung nha

câu 2:

bạn chuyển xy sang  vế trái rồi nhóm với x hoặc y nha, cái còn lại thì bạn nhóm với 1 và cũng đặt nhân tử chung sau đó thì bạn tính ra nha

BẠN MÀ K LÀM ĐC THÌ CHỊU ĐÓ :)))

mai thùy trang ví dụ mà đưa xy sang vế trái thì sẽ đc là x +y+1 -xy=0 thì là đc x(y-1)+(y+1) hoặc là y(x-1)+(x+1) chứ lm j mà nhóm nhân tử chung đk bn

mai huy thang

Bài 1:

|x-2|=4-x

ĐK: \(4-x\ge0\Leftrightarrow x\le4\)

Ta có: \(\orbr{\begin{cases}x-2=4-x\\x-2=x-4\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\0=2\left(loại\right)\end{cases}\Rightarrow}}x=3\left(tm\right)\)

Vậy x = 3 

Bài 2:

a, sao có z

b, Vì \(\hept{\begin{cases}\left|2017-x\right|\ge0\\\left|y-x+2018\right|\ge0\end{cases}\Rightarrow\left|2017-x\right|+\left|y-x+2018\right|\ge0}\)

Mà |2017-x|+|y-x+2018|=0

\(\Rightarrow\hept{\begin{cases}\left|2017-x\right|=0\\\left|y-x+2018\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\y-2017+2018=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2017\\y=1\end{cases}}}\)

Vậy x=2017,y=1

c, giống b

Bài 2 cũng có z bạn ạ Làm luôn hộ mình câu b

b) ta thấy /2017-x/>=0

/y-x+2018/>= 0

=> /2017-x/+/y-x+2018/>=0

dấu = xảy ra khi 2017-x=0 => x=2017

                     và y-x+2018=0 => y= 1

vậy (x;y)=(2017;1)