1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)

Vậy ...................

b/ ĐKXĐ:\(x\ne2;x\ne5\)

.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x^2-10x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)

Vậy ..............

`Answer:`

`1.`

a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)

b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)

\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)

`2.`

\(ĐKXĐ:x\ne-m-2;x\ne m-2\)

Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)

a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)

b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì

\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)

Đặt \(x+\dfrac{1}{x}=a;y+\dfrac{1}{y}=b\left(\left|a\right|\ge2;\left|b\right|\ge2\right)\)

\(\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x^3+\dfrac{1}{x^3}\right)+\left(y^3+\dfrac{1}{y^3}\right)=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)^3-3\left(y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\a^3+b^3=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(a+b\right)^3-3ab\left(a+b\right)=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\125-15ab=15m-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=9-m\end{matrix}\right.\)

\(\Rightarrow a,b\) là nghiệm của phương trình \(t^2-5t+9-m=0\left(1\right)\)

a, Nếu \(m=3\), phương trình \(\left(1\right)\) trở thành

\(t^2-5t+6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\)

TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y+\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\y^2-3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\pm\sqrt{5}}{2}\end{matrix}\right.\)

TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=3\\y+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3\pm\sqrt{5}}{2}\\y=1\end{matrix}\right.\)

Vậy ...

b, \(\left(1\right)\Leftrightarrow t=\dfrac{5\pm\sqrt{4m-11}}{2}\left(m\ge\dfrac{11}{4}\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\pm\sqrt{4m-11}}{2}\\b=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5\pm\sqrt{4m-11}}{2}\\y+\dfrac{1}{y}=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-\left(5\pm\sqrt{4m-11}\right)+2=0\left(2\right)\\2y^2-\left(5\mp\sqrt{4m-11}\right)+2=0\end{matrix}\right.\)

Yêu cầu bài toán thỏa mãn khi phương trình \(\left(2\right)\) có nghiệm dương

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(5\pm\sqrt{4m-11}\right)^2-16\ge0\\\dfrac{5\pm\sqrt{4m-11}}{2}>0\\1>0\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(\Delta'=\left[-\left(m+1\right)\right]^2-\left(2m-2\right)\)

= m² + 2m + 1 - 2m + 2 = m² + 3 > 0 (vì m² ≥ 0)

⇒ Phương trình có 2 nghiệm phân biệt x₁, x₂

Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m-2\end{matrix}\right.\)

Ta có: x₁² + x₂² + 3x₁x₂ = 25

⇔ (x₁ + x₂)² - 2x₁x₂ + 3x₁x₂ = 25

⇔ (x₁ + x₂)² + x₁x₂ = 25

⇔ [2(m + 1)]² + (2m - 2) = 25

⇔ 4m² + 8m + 4 + 2m - 2 - 25 = 0

⇔ 4m² + 10m - 23 = 0

⇔ \(\left[{}\begin{matrix}m=\dfrac{-5+3\sqrt{13}}{4}\\m=\dfrac{-5-3\sqrt{13}}{4}\end{matrix}\right.\)

Vậy m = ...

Thay `m=-3` ta có:

`(x+3)/(x+5)+(x-5)/(x-3)=2`

`<=>(x^2-9+x^2-25)/((x+5)(x-3))=2`

`<=>(2x^2-34)/(x^2+2x-15)=2`

`<=>2x^2-34=2x^2+4x-30`

`<=>4x=-4`

`<=>x=-1`

Vậy `S={-1}`

a) Với m = 5 ta có pt: (100 - 25)x - 5 = 10

<=> 75 x = 15 <=> x = 1/5

b) (4m² - 25) - 5 = 2m

<=> 4m² - 2m - 30 = 0

<=> 4m² + 10m - 12m - 30 = 0

<=> (m - 3)(4m + 10) = 0 <=> \(\orbr{\begin{cases}m=3\\m=-\frac{5}{2}\end{cases}}\)

a) với m = 5, ta có:

(4.5² - 25)x - 5 = 2.5

<=> (100 - 25)x - 5 = 10

<=> 75x - 5 = 10

<=> 75x = 10 + 5

<=> 75x = 15

<=> x = 15/75 = 1/5

b) (1.4m² - 25).1 - 5 = 2.m 

<=> (4m² - 25) - 5 = 2m

<=> 4m² - 25 - 5 = 2m

<=> 4m² - 30 = 2m

<=> 4m² - 30 - 2m = 0

<=> 2(2m² - 15 - m) = 0

<=> 2(2m² + 5m - 6m - 15) = 0

<=> 2[m(2m + 5) - 3(2m + 5)] = 0

<=> 2(2m + 5)(m - 3) = 0

<=> 2m + 5 = 0 hoặc m - 3 = 0

<=> m = -5/2 hoặc m = 3

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-5\\x_1x_2=-3m\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x_3=\dfrac{2}{x_1^2}\\x_4=\dfrac{2}{x^2_2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_3+x_4=\dfrac{2}{x_1^2}+\dfrac{2}{x_2^2}\\x_3x_4=\dfrac{4}{x_1^2x_2^2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_3+x_4=\dfrac{2\left(x_1+x_2\right)^2-4x_1x_2}{\left(x_1x_2\right)^2}\\x_3x_4=\dfrac{4}{\left(x_1x_2\right)^2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_3+x_4=\dfrac{2.\left(-5\right)^2-4\left(-3m\right)}{\left(-3m\right)^2}=\dfrac{12m+50}{9m^2}\\x_3x_4=\dfrac{4}{\left(-3m\right)^2}=\dfrac{4}{9m^2}\end{matrix}\right.\)

\(\Rightarrow x_3;x_4\) là nghiệm:

\(x^2-\left(\dfrac{12m+50}{9m^2}\right)x+\dfrac{4}{9m^2}=0\)

\(\Leftrightarrow9m^2x^2-\left(12m+50\right)x+4=0\)

Có vô só nghiệm khi 0x +0 = 0 vậy phải thỏa mãm hai đk :

m  - 5 = 0      => m = 5  

m^2 - 5 = 0  => ( m - 5) ( m+5) = 0 => m = 5 hoặc -5

Kết hợp hai đk => m = 5 thì pt vô số nghiệm

Tham khảo bài này :

(3x+1)(7x+3)=(5x-7)(3x+1)

<=> (3x+1)(7x+3)-(5x-7)(3x+1)=0

<=> (3x+1)(7x+3-5x+7)=0

<=> (3x+1)(2x+10)=0

<=> 2(3x+1)(x+5)=0

=> 3x+1=0 hoặc x+5=0

=> x= -1/3 hoặc x=-5

Vậy x = -1/3 hoặc x = -5

\(a,x^2+10x+25-4x\left(x+5\right)=0.\)

\(\Leftrightarrow\left(x+5\right)^2-4x\left(x+5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\5-3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}}\)

\(b,\left(4x-5\right)^2-2\left(16x^2-25\right)=0\)

\(\Leftrightarrow\left(4x-5\right)^2-2\left(4x+5\right)\left(4x-5\right)=0\)

\(\Leftrightarrow-\left(4x-5\right)\left(4x+15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x-5=0\\4x+15=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{4}\\x=-\frac{15}{4}\end{cases}}}\)

Câu 1: 

\(sin^2x+cos^2x=1\Leftrightarrow cos^2x=1-sin^2x=1-\left(\frac{4}{5}\right)^2=\frac{9}{15}\)

\(\Leftrightarrow cos^2x=\frac{\pm3}{5}\).

Câu 2: 

Đường tròn \(\left(C\right)\)có tâm \(I\left(2,-1\right)\)bán kính \(R=\sqrt{25}=5\).

Gọi \(d\)là tiếp tuyến của đường tròn \(\left(C\right)\)tại điểm \(M\). Khi đó \(IM\)và \(d\)vuông góc với nhau.

\(\Rightarrow\overrightarrow{IM}=\left(3,4\right)\)là một vector pháp tuyến của \(d\)

Suy ra phương trình \(d:3\left(x-5\right)+4\left(y-3\right)=0\Leftrightarrow3x+4y-27=0\).