\(\left(\dfrac{2x}{3}-\dfrac{1}{3}\right)+\left(3x-2x+1\right)=8\)

\(\Leftrightarrow\dfrac{2x-1}{3}+x-7=0\Rightarrow2x-1+3x-21=0\Leftrightarrow x=\dfrac{22}{5}\)

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left[3x-2\left(x-1\right)\right]=8\)

\(\Rightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+2=8\)

\(\Rightarrow\dfrac{5}{3}x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{5}\)

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)+\left(3x-2\left(x-1\right)\right)=8\)

\(\Leftrightarrow\dfrac{2}{3}x-\dfrac{1}{3}+3x-2x+1=8\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{22}{3}\)

\(\Leftrightarrow x=\dfrac{22}{5}\)

\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{16}\) và \(x^2-2y^2+z^2=8\)

Áp dụng t/c dãy tsbn:

\(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{2y^2}{18}=\dfrac{z^2}{16}=\dfrac{x^2-2y^2+z^2}{4-18+16}=\dfrac{8}{2}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=16\\y^2=36\\z^2=64\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\pm4\\y=\pm6\\z=\pm8\end{matrix}\right.\)

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)

Ta có: \(x^2-2y^2+z^2=8\)

\(\Leftrightarrow4k^2-18k^2+16k^2=8\)

\(\Leftrightarrow k^2=4\)

Trường hợp 1: k=2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=4k=8\end{matrix}\right.\)

Trường hợp 2: k=-2

\(\Leftrightarrow\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=4k=-8\end{matrix}\right.\)

a: \(\Leftrightarrow3x+9=-2x+6\)

=>5x=-3

hay x=-3/5

b: =>3/x=y/35=3/7

=>x=7; y=15

c: =>9x/5=-3/5

=>9x=-3

hay x=-1/3

d: =>x+2/26=-1/4

=>x+2=-13/2

hay x=-17/2

\(\dfrac{7}{9}\times\dfrac{3}{14}-\dfrac{5}{8}=\dfrac{21}{126}-\dfrac{5}{8}=\dfrac{1}{6}-\dfrac{5}{8}=-\dfrac{11}{24}\)

= 1/6 - 5/8

=-11/24

7/9 x 3/14 - 5/8

= 1/6       -  5/8

=      -11/24

........

a, (1) có nghiệm duy nhất trên [-2 ; 2] khi

[-2 ; 2] khi \(\left[{}\begin{matrix}-4m=-8\\1\ge-4m>-7\end{matrix}\right.\)

⇔ \(\left[{}\begin{matrix}m=2\\\dfrac{-1}{4}\le m< \dfrac{7}{4}\end{matrix}\right.\) hay m ϵ [\(\dfrac{-1}{4};\dfrac{7}{4}\)) \(\cup\left\{2\right\}\)

(1) có nghiệm duy nhất trên [2 ; 3] khi

- 4 ≥ - 4m ≥ - 7 ⇔ 1 ≤ m ≤ \(\dfrac{7}{4}\) hay m ∈\(\left[1;\dfrac{7}{4}\right]\)

(1) có nghiệm duy nhất trên  [-2; -1] khi 

-4 ≤ 4m ≤ 1 hay m ∈ \(\left[\dfrac{-1}{4};1\right]\)

b, (1) có 2 nghiệm phân biệt trên [-2 ; 2] khi

-4m ∈ (-8 ; -7] ⇒ m ∈\(\)[\(\dfrac{7}{4}\); 2)

(1) có 2 nghiệm phân biệt trên [2; 3] và [-2; -1] khi m ∈ ∅

c, (1) có nghiệm trên đoạn 

[-2; 2] khi -8 ≤ -4m ≤ 1 ⇒ m ∈ \(\left[\dfrac{-1}{4};2\right]\)

[2 ; 3] khi - 4 ≥ - 4m ≥ - 7  hay m ∈\(\left[1;\dfrac{7}{4}\right]\)

[-2 ; -1] khi -4 ≤ 4m ≤ 1 hay m ∈ \(\left[\dfrac{-1}{4};1\right]\)

d, dường như là nó giống câu b,

e, (1) vô nghiệm trên đoạn [-2 ; 2] khi 

\(\left[{}\begin{matrix}-4m>1\\-4m< -8\end{matrix}\right.\)hay \(m\in\left(-\infty;\dfrac{-1}{4}\right)\cup\left(2;+\infty\right)\)

(1) vô nghiệm trên đoạn [2; 3] khi 

m ∈ R \ \(\left[1;\dfrac{7}{4}\right]\)

(1) vô nghiệm trên [-2 ; -1] khi m ∈ R \ \(\left[\dfrac{-1}{4};1\right]\)

Có sai sót xin thông cảm

P/s :Bạn tự vẽ bảng biến thiên nha, nhớ chia khoảng cách các giá trị của x cho chuẩn vào, nhớ thêm cả f(0) và trong bảng nhá

Đề sai rồi bạn

kệ bn :v

ĐK : \(x\ne-2.-3;-4;-5;-6\)

\(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\Leftrightarrow\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\Leftrightarrow x^2+8x-20=0\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\Leftrightarrow x=2;x=-10\)( tmđkxđ )

Vậy tập nghiệm phương trình là S = { -10 ; 2 } 

ĐKXĐ \(x\notin\left\{-2;-3;...;-6\right\}\)

Phương trình tương đương với:

\(\dfrac{1}{\left(x^2+2x\right)+\left(3x+6\right)}+\dfrac{1}{\left(x^2+3x\right)+\left(4x+12\right)}+\dfrac{1}{\left(x^2+4x\right)+\left(5x+20\right)}+\dfrac{1}{\left(x^2+5x\right)+\left(6x+30\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{\left(x+3\right)-\left(x+2\right)}{\left(x+2\right)\left(x+3\right)}+\dfrac{\left(x+4\right)-\left(x+3\right)}{\left(x+3\right)\left(x+4\right)}+\dfrac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\dfrac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\\ \Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x+2\right)\left(x+6\right)}=\dfrac{4}{32}\\ \Rightarrow\left(x+2\right)\left(x+6\right)=32\\\Leftrightarrow x^2+8x-20=0\\ \Leftrightarrow\left(x+10\right)\left(x-2\right)=0\\ \Leftrightarrow\begin{matrix}x=2\\x=-10\end{matrix}\left(t.m\right)\)

ĐKXĐ: \(x\notin\left\{-2;-3;-4;-5;-6\right\}\)

Ta có: \(\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}+\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{8\left(x+6\right)}{8\left(x+2\right)\left(x+6\right)}-\dfrac{8\left(x+2\right)}{8\left(x+2\right)\left(x+6\right)}=\dfrac{\left(x+2\right)\left(x+6\right)}{8\left(x+2\right)\left(x+6\right)}\)

Suy ra: \(8x+48-8x-16=x^2+8x+12\)

\(\Leftrightarrow x^2+8x+12-32=0\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow x^2+10x-2x-20=0\)

\(\Leftrightarrow x\left(x+10\right)-2\left(x+10\right)=0\)

\(\Leftrightarrow\left(x+10\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

Vậy: S={-10;2}

\(\sqrt{2x+7m}=x+2\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x+7m=x^2+4x+4\\x+2\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+2x+4=7m\left(1\right)\\x\ge-2\end{matrix}\right.\)

Yêu cầu bài toán thỏa mãn khi phương trình \(\left(1\right)\) có nghiệm \(x\ge-2\)

Bảng biến thiên:

Dựa vào bảng biến thiên ta được \(7m\ge3\Leftrightarrow m\ge\dfrac{3}{7}\)