a) Dễ thấy Ix+1I +Ix+2I+ Ix+3I >= 0 nên 4x >=0 \(\Rightarrow\)x>= 0

 Suy ra 4x=x+1+x+2+x+3= 3x+6 , x=6

Các phần khác tương tự

Ta có: \(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|\ge0\left(\forall x\right)\)

=> \(5x\ge0\left(\forall x\right)\)

<=> \(x\ge0\left(\forall x\right)\)

Thay vào ta được:

\(x+1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow4x+5=5x\)

\(\Rightarrow x=5\)

Ta có:  |x+1,1|\(\ge\)0

            |x+1,2|\(\ge\)0

            |x+1,3|\(\ge\)0

            |x+1,4|\(\ge\)0

Suy ra:  |x+1,1|+|x+1,2|+|x+1,3|+|x+1,4|\(\ge\)0

   <=>    5x\(\ge\)0

     =>      x\(\ge\)0

Do đó: |x+1,1|+|x+1,2|+|x+1,3|+|x+1,4|=5x

  <=>           x+1,1+x+1,2+x+1,3+x+1,4=5x

                           4x+(1,1+1,2+1,3+1,4)=5x

                                           4x+5           =5x

                                           4x               =5x-5

                                           4x-5x          =-5

                                           (4-5)x          =-5

                                                -1x          =-5

   =>                                           1x         =5

                                                    x          =5:1

   =>                                             x          =5

Vậy x cần tìm là 5

1, https://hoc24.vn//hoi-dap/question/91350.html

Bài 3:

=>2xy-x-y-2=0

=>x(2y-1)-y+0,5-2,5=0

=>x(2y-1)-(y-0,5)=2,5

=>2x(2y-1)-(2y-1)=5

=>(2y-1)(2x-1)=5

=>\(\left(2x-1;2y-1\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(0;-2\right);\left(-2;0\right)\right\}\)

Câu 5:

Đặt x/2=y/3=z/4=k

=>x=2k; y=3k; z=4k

x^2+y^2+z^2=116

=>4k^2+9k^2+16k^2=116

=>29k^2=116

=>k^2=4

TH1: k=2

=>x=4; y=6; z=8

TH2: k=-2

=>x=-4; y=-6; z=-8

a) \(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7};x+y+z=56\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{56}{14}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.2=8\\y=4.5=20\\z=4.7=28\end{matrix}\right.\)

b) \(\dfrac{x}{1,1}=\dfrac{y}{1,3}=\dfrac{z}{1,4}\left(1\right);2x-y=5,5\)

\(\left(1\right)\Rightarrow\dfrac{2x-y}{1,1.2-1,3}=\dfrac{5,5}{0,9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=1,1.\dfrac{5,5}{0,9}=\dfrac{6,05}{0,9}\\y=1,3.\dfrac{5,5}{0,9}=\dfrac{7,15}{0,9}\\z=\dfrac{1,4}{1,1}.x=\dfrac{1,4}{1,1}.\dfrac{6,05}{0,9}=\dfrac{8,47}{0,99}\end{matrix}\right.\)

d) \(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5};xyz=-30\)

\(\dfrac{x}{2}=\dfrac{x}{3}=\dfrac{z}{5}=\dfrac{xyz}{2.3.5}=\dfrac{-30}{30}=-1\)

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

a) �2=�5=�7;�+�+�=562x​=5y​=7z​;x+y+z=56

�2=�5=�7=�+�+�2+5+7=5614=42x​=5y​=7z​=2+5+7x+y+z​=1456​=4

⇒{�=4.2=8�=4.5=20�=4.7=28⇒⎩⎨⎧​x=4.2=8y=4.5=20z=4.7=28​

b) �1,1=�1,3=�1,4(1);2�−�=5,51,1x​=1,3y​=1,4z​(1);2x−y=5,5

(1)⇒2�−�1,1.2−1,3=5,50,9(1)⇒1,1.2−1,32x−y​=0,95,5​
    

⇒⎩⎨⎧​x=1,1.0,95,5​=0,96,05​y=1,3.0,95,5​=0,97,15​z=1,11,4​.x=1,11,4​.0,96,05​=0,998,47​​

d) �2=�3=�5;���=−302x​=3x​=5z​;xyz=−30

�2=�3=�5=���2.3.5=−3030=−12x​=3x​=5z​=2.3.5xyz​=30−30​=−1

⇒{�=2.(−1)=−2�=3.(−1)=−3�=5.(−1)=−5⇒⎩⎨⎧​x=2.(−1)=−2y=3.(−1)=−3z=5.(−1)=−5​
 

chết rùi nó bị lỗi

xin lỗi nha

\(\left|x+1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)(1)

VT(1) >=0 với mọi x nên để 1 có nghiệm thì 5x phải >= 0 hay x>=0

Với x>=0 thì các giá trị tuyệt đối của VT bằng biểu thức bên trong nên

(1) <=> x + 1,1 + x + 1,2 + x + 1,3 + x + 1,4 = 5x

<=> x = 5.