Đặt\(A=\left|x+1\right|+\left|3-x\right|=7\)

\(x+1=0\Leftrightarrow x=-1\)

\(3-x=0\Leftrightarrow x=3\)

\(+x< -1\)

\(\Rightarrow-\left(x+1\right)+3-x=7\)

\(2-2x=7\)

\(2x=-5\)

\(x=-\frac{5}{2}\left(tm\right)\)

\(+:-1\le x\le3\)

\(x+1-3+x=7\)

\(2x-2=7\)

\(x=\frac{9}{2}\)

\(+:x\ge3\)

\(x+1+3-x=7\left(l\right)\)

Vậy...........

mik ở đâu thì bảo mik vs

cảm ơn bạn nhiều nhé

Ta có: \(\hept{\begin{cases}|x+2y-z|\ge0;\forall x,y,z\\\left(x-y+3z\right)^2\ge0;\forall x,y,z\\\left(z-1\right)^4\ge0;\forall x,y,z\end{cases}}\)\(\Rightarrow|x+2y-z|+\left(x-y+3z\right)^2+\left(z-1\right)^4\ge0;\forall x,y,z\)

Do đó \(|x+2y-z|+\left(x-y+3z\right)^2+\left(z-1\right)^4=0\)

\(\Leftrightarrow\hept{\begin{cases}|x+2y-z|=0\\\left(x-y+3z\right)^2=0\\\left(z-1\right)^4=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+2y-z=0\\x-y+3z=0\\z=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+2y=1\\x-y=-3\\z=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-5}{3}\\y=\frac{4}{3}\\z=1\end{cases}}\)

Vậy ...

Mình cần gấp ạ 

Ta có: \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}=\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)

                                              \(=\frac{2^{12}.\left(5^{10}-5^8\right)}{2^{12}.5^8-2^{14}.5^8}\)

                                             \(=\frac{2^{12}.5^8.\left(5^2-1\right)}{5^8.\left(2^{12}-2^{14}\right)}\)

                                             \(=\frac{2^{12}.5^8.24}{5^8.2^{12}.\left(1-2^2\right)}\)

                                             \(=\frac{24}{-3}\)

                                              \(=-8\)

Học "tuốt" nha^^

(-3/4 + 2/3) : 5/11 + (-1/4 + 1/3) : 5/11

= [ (-3/4 + 2/3) + ( -1/4 + 1/3 ) : 5/11

= [ ( -9/12 + 8/12 ) + ( -3/12 + 4/12) .11/5

= [ ( -1/12) + 1/12 ] .11/5

= 0. 11/5

= 0

Ok , mình sẽ làm !

Ta có :

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b}{c}-1+1=\frac{b+c}{a}-1+1=\frac{c+a}{b}-1+1\)

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\left(1\right)\)

+) Trường hợp 1 : \(a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

Ta có :

 \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-a}{a}.\frac{-c}{c}.\frac{-b}{b}\)

\(\Leftrightarrow P=-1.\left(-1\right).\left(-1\right)=-1\)

+) Trường hợp 2 : \(a+b+c\ne0\)

Áp dụng tính chất của dãy tỉ số bằng nhau cho ( 1 ) , ta có :

\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)

Ta lại có :

 \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

\(\Leftrightarrow P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{c+b}{b}\)

\(\Leftrightarrow P=2.2.2=8\)

Vậy....................

Đề sai nhé bạn ! Bạn kiểm tra lại!