Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=\pm\sqrt{\frac12}\end{array}\right.\)

Vậy \(x\) ∈ {- \(\sqrt{\frac12}\); 0; \(\sqrt{\frac12}\)}

Giải:

\(x-5\sqrt{x}\) = 0 (\(x\) ≥ 0)

\(\sqrt{x}\) .(\(\sqrt{x}\) - 5) = 0

\(\left[\begin{array}{l}\sqrt{x}=0\\ \sqrt{x}-5=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ \sqrt{x}=5\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=25\end{array}\right.\)

Vậy \(x\in\) {0; 25}

\(x^5\) = 2\(x^7\)

\(x^5\) - 2\(x^7\) = 0

\(x^5\).(1 - 2\(x^2\)) = 0

\(\left[\begin{array}{l}x^5=0\\ 1-2x^2=0\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ 2x^2=1\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x^2=\frac12\end{array}\right.\)

\(\left[\begin{array}{l}x=0\\ x=-\frac{1}{\sqrt2}\\ x=\frac{1}{\sqrt2}\end{array}\right.\)

Vậy \(x\) \(\in\) {- \(\frac{1}{\sqrt2}\); 0; \(\frac{1}{\sqrt2}\)}

a)\(\left(x+1\right)\left(x-5\right)< 0\) khi \(\left(x+1\right)\) và \(\left(x-5\right)\) trái dấu.

Chú ý rằng: \(x+1>x-5\) nên \(x+1>0,x-5< 0\). Giải cả hai trường hợp ta có:

\(\hept{\begin{cases}x+1>0\\x-5< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-1\\x< 5\end{cases}}\Leftrightarrow-1< x< 5}\)

b) \(\left(x-2\right)\left(x+\frac{5}{7}\right)>0\) khi \(\left(x-2\right)\) và \(\left(x+\frac{5}{7}\right)\) đồng dấu (\(x-2\ne0,\left(x+\frac{5}{7}\right)\ne0\Leftrightarrow x\ne2;x\ne-\frac{5}{7}\)

+ Với \(\left(x-2\right)\) và \(\left(x+\frac{5}{7}\right)\) dương thì ta có:\(x-2< x+\frac{5}{7}\). Có 2 TH

 \(\hept{\begin{cases}x-2>0\\x+\frac{5}{7}>0\end{cases}\Leftrightarrow\hept{\begin{cases}x>2\\x>-\frac{5}{7}\end{cases}}}\) . Dễ thấy để thỏa mãn cả hai trường hợp thì x > 2  (1)

+ Với \(\left(x-2\right)\) và \(\left(x+\frac{5}{7}\right)\) âm thì ta có: \(x-2< x+\frac{5}{7}\). Có 2 TH

\(\hept{\begin{cases}\left(x-2\right)< 0\\\left(x+\frac{5}{7}\right)< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 2\\x< -\frac{5}{7}\end{cases}}}\). Dễ thấy để x thỏa mãn cả hai trường hợp thì \(x< -\frac{5}{7}\) (2)

Từ (1) và (2) ta có: \(\hept{\begin{cases}x>2\\x< -\frac{5}{7}\end{cases}}\) thì \(\left(x-2\right)\left(x+\frac{5}{7}\right)>0\)

1. \(\frac{x+1}{x+5}=\frac{x+3}{x+2}\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)=\left(x+3\right)\left(x+5\right)\)

\(\Leftrightarrow\left(x+1\right)x+\left(x+1\right).2=\left(x+3\right)x+\left(x+3\right).5\)

\(\Leftrightarrow x^2+x+2x+2=x^2+3x+5x+15\)

\(\Leftrightarrow x^2+3x+2=x^2+8x+15\)

\(\Leftrightarrow x^2+3x-x^2-8x=15-2\)

\(\Leftrightarrow-5x=13\)

\(\Leftrightarrow x=\frac{-13}{5}\)

Vậy ...

Cảm ơn bn nha 

a)Vì  \(-|x-3,5|\le0;\forall x\)

\(\Rightarrow0,5-|x-3,5|\le0,5-0;\forall x\)

Hay \(A\le0,5-0;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-3,5=0\)

                        \(\Leftrightarrow x=3,5\)

Vậy MAX A=0,5 \(\Leftrightarrow x=3,5\)

b) Vì \(-|1,4-x|\le0;\forall x\)

\(\Rightarrow-|1,4-x|-2\le0-2;\forall x\)

Hay \(B\le-2;\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow1,4-x=0\)

                        \(\Leftrightarrow x=1,4\)

Vậy MAX B=-2 \(\Leftrightarrow x=1,4\)

1: xy+x+y+1=0

=>x(y+1)+(y+1)=0

=>(x+1)(y+1)=0

=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)

2: xy+x+6=0

=>x(y+1)=-6

=>(x;y+1)∈{(1;-6);(-6;1);(-1;6);(6;-1);(2;-3);(-3;2);(-2;3);(3;-2)}

=>(x;y)∈{(1;-7);(-6;0);(-1;5);(6;-2);(2;-4);(-3;1);(-2;2);(3;-3)}

3: -xy-x-y-1=0

=>xy+x+y+1=0

=>x(y+1)+(y+1)=0

=>(x+1)(y+1)=0

=>\(\begin{cases}x+1=0\\ y+1=0\end{cases}\Rightarrow\begin{cases}x=-1\\ y=-1\end{cases}\)

4: xy-x-y+1=0

=>x(y-1)-(y-1)=0

=>(x-1)(y-1)=0

=>\(\begin{cases}x-1=0\\ y-1=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=1\end{cases}\)

5: xy+2x+y+11=0

=>x(y+2)+y+2+9=0

=>x(y+2)+(y+2)=-9

=>(x+1)(y+2)=-9

=>(x+1;y+2)∈{(1;-9);(-9;1);(-1;9);(9;-1);(3;-3);(-3;3)}

=>(x;y)∈{(0;-11);(-10;-1);(-2;7);(8;-3);(2;-5);(-4;1)}

6: ĐKXĐ: x<>0

\(\frac{5}{x}+\frac{y}{4}=\frac18\)

=>\(\frac{20+xy}{4x}=\frac18\)

=>\(\frac{40+2xy}{8x}=\frac{x}{8x}\)

=>40+2xy=x

=>x-2xy=40

=>x(1-2y)=40

=>x(2y-1)=-40

mà 2y-1 lẻ(do y nguyên)

nên (x;2y-1)∈{(-40;1);(40;-1);(8;-5);(-8;5)}

=>(x;2y)∈{(-40;2);(40;0);(8;-4);(-8;6)}

=>(x;y)∈{(-40;1);(40;0);(8;-2);(-8;3)}

8: (x+2)(y-3)=-3

=>(x+2;y-3)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(-1;0);(-5;4);(-3;6);(1;2)}

Các bạn giúp mik giải bài 1 thôi cũng được vì bài 2 mik lm đc rồi. Hiện tại mik đag cần gấp giúp mik với

\(20^x\div14^x=\left(\frac{20}{14}\right)^x=\left(\frac{10}{7}\right)^x=\frac{10}{7}x\)

Mình biết kết quả là x = 1 nhưng chưa tìm ra cách giải, bạn cố gắng tìm nốt nhé!

Bài 1:

a)  \(x-\frac{20}{11.13}-\frac{20}{13.15}-...-\frac{20}{53.55}=\frac{3}{11}\)

\(x-\left(\frac{20}{11.13}+\frac{20}{13.15}+...+\frac{20}{53.55}\right)=\frac{3}{11}\)

\(x-\frac{20}{2}.\left(\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+...+\frac{1}{53}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10.\left(\frac{1}{11}-\frac{1}{55}\right)=\frac{3}{11}\)

\(x-10\cdot\frac{4}{55}=\frac{3}{11}\)

\(x-\frac{8}{11}=\frac{3}{11}\)

\(x=1\)

b) \(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x.\left(x+1\right)}=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(2.\left(\frac{1}{6}-\frac{1}{x+1}\right)=\frac{2}{9}\)

\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\frac{1}{x+1}=\frac{1}{18}\)

=> x + 1 =18

x = 17

bài 2 ko bk lm, xl nha

mk cảm ơn bn nha

Ta có : \(\hept{\begin{cases}\left|5-\frac{2}{3}x\right|\ge0\forall x\\\left|\frac{1}{7}y-3\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|5-\frac{2}{3}x\right|+\left|\frac{1}{7}y-3\right|\ge0\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}5-\frac{2}{3}x=0\\\frac{1}{7}y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\y=21\end{cases}}\)

b) Ta có \(\hept{\begin{cases}\left|5x+10\right|\ge0\forall x\\\left|6y-9\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|5x+10\right|+\left|6y-9\right|\ge0\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}5x+10=0\\6y-9=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1,5\end{cases}}\)