\(a.\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)

\(\Leftrightarrow\left(x^2+7x-4x+16-14\right)\left(x^2+7x+4x+16+14\right)-60=0\)

\(\Leftrightarrow\left(x^2+7x+16-4x-14\right)\left(x^2+7x+16+4x+14\right)=0\)

\(\Leftrightarrow\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60=0\)

Vì \(\left(x^2+7x+16\right)^2>0;\left(4x+14\right)^2>0\)

Nên \(\left(x^2+7x+16\right)^2-\left(4x+14\right)^2-60\ge-60\)

V...\(S=\varnothing\)

\(b.4^x-12.2^x+32=0\)

\(\Leftrightarrow\left(2^x\right)^2-2.2^x.6+36-4=0\)

\(\Leftrightarrow\left(2^x-6\right)^2-4=0\)

\(\Leftrightarrow\left(2^x-4\right)\left(2^x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2^x-4=0\\2^x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2^x=4\\2^x=8\end{cases}\Leftrightarrow}\orbr{\begin{cases}2^x=2^2\\2^x=2^3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=3\end{cases}}}\)

V...\(S=\left\{2;3\right\}\)

^^ đúng ko ta

a) (x+1)(x+2)(x+5)(x+6)-60=0

[(x+1)(x+6)][(x+2)(x+5)]-60=0

(x^2 + 7x + 6)(x^2  + 7x + 10) - 60 = 0

đặt t = x^2 + 7x + 8

pt trở thành

(t-2)(t+2)-60=0

t^2 - 64=0 .....

t=8 hoặc t=-8.

tìm x ....

a) Từ phương trình ban đầu ta có:

(x + 1)(x + 2)(x + 5)(x + 6) = 60

\(\Leftrightarrow\) [(x + 1)(x + 6)][(x + 2)(x + 5)] = 60

\(\Leftrightarrow\) (x² + 7x + 6)(x² + 7x + 10) = 60 (1)

Đặt x² + 7x + 6 = a. Thay a vào phương trình (1) ta có:

a(a + 4) = 60

\(\Leftrightarrow\) a² + 4a + 4 = 64

\(\Leftrightarrow\) (a + 2)² = 64

\(\Leftrightarrow\) a + 2 = \(\pm\)8

Đến đây thay x vào rồi giải tiếp

Cái Này bạn bấm máy tinh nha

Bạn Ghi Cái đề bài vào Xong bấm SHIFT  rồi  Bấm CALC rồi Bấm = 

Là Ra Nhé Nhớ Cho mình Nha

     \(\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)

\(\Rightarrow\left[\left(x+1\right)\left(x+2\right)\right].\left[\left(x+5\right)\left(x+6\right)\right]-60=0\)

\(\Rightarrow\left[\left(x+1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+5\right)\right]-60=0\)

\(\Rightarrow\left(x^2+7x+6\right)\left(x^2+7x+10\right)-60=0\left(1\right)\)

Đặt \(x^2+7x+6=a\Rightarrow x^2+7x+10=a+4\)

Thay vào (1), ta có:

     \(a\left(a+4\right)-60=0\)

\(\Rightarrow a^2+4a-60=0\)

\(\Rightarrow a^2+10a-6a-60=0\)

\(\Rightarrow a\left(a+10\right)-6\left(a+10\right)=0\)

\(\Rightarrow\left(a-6\right)\left(a+10\right)=0\)

\(\Rightarrow\orbr{\begin{cases}a=6\\a=-10\end{cases}}\)

- Nếu \(x^2+7x+6=6\)

\(\Rightarrow x^2+7x=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)

- Nếu \(x^2+7x+6=-10\)

\(\Rightarrow x^2+7x+16=0\)

Mà \(x^2+7x+16=x^2+2.x.\frac{7}{2}+\frac{49}{4}+\frac{15}{4}=\left(x+\frac{7}{2}\right)^2+\frac{15}{4}>0\forall x\)

Vậy \(x=0,x=-7\)

Học tốt.

\(\left(3x-2\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

ĐKXĐ: x khác -4;-5;-6;-7

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right).\left(x+5\right)}+\frac{1}{\left(x+5\right).\left(x+6\right)}+\frac{1}{\left(x+6\right).\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\frac{1}{18}\Rightarrow3.18=x^2+11x+28\)

\(\Rightarrow x^2+11x-26=0\)

\(\Rightarrow\left(x-2\right).\left(x+13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}\left(tm\right)}\)

Vậy...

c.

ĐKXĐ: \(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

\(\Leftrightarrow x+4-2\sqrt[]{\left(\dfrac{x+2}{x-1}\right)^2\left(\dfrac{x-1}{x+2}\right)}=0\)

\(\Leftrightarrow x+4-2\sqrt[]{\dfrac{x+2}{x-1}}=0\)

\(\Leftrightarrow x+4=2\sqrt[]{\dfrac{x+2}{x-1}}\) (\(x\ge-4\))

\(\Leftrightarrow x^2+8x+16=\dfrac{4\left(x+2\right)}{x-1}\)

\(\Rightarrow x^3+7x^2+4x-24=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+4x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2+2\sqrt{3}\\x=-2-2\sqrt{3}\left(loại\right)\end{matrix}\right.\)

a.

\(\Leftrightarrow2x^2-11x+21=3\sqrt[3]{4\left(x-1\right)}\)

Do \(2x^2-11x+21=2\left(x-\dfrac{11}{4}\right)^2+\dfrac{47}{8}>0\Rightarrow3\sqrt[3]{4\left(x-1\right)}>0\Rightarrow x-1>0\)

Ta có:

\(VT=2x^2-11x+21-3\sqrt[3]{4x-4}=2\left(x^2-6x+9\right)+x+3-3\sqrt[3]{4\left(x-1\right)}\)

\(=2\left(x-3\right)^2+x+3-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge x+3-3\sqrt[3]{4\left(x-1\right)}=\left(x-1\right)+2+2-3\sqrt[3]{4\left(x-1\right)}\)

\(\Rightarrow VT\ge3\sqrt[3]{\left(x-1\right).2.2}-3\sqrt[3]{4\left(x-1\right)}=0\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\\x-1=2\\\end{matrix}\right.\) \(\Leftrightarrow x=3\)

Vậy pt có nghiệm duy nhất \(x=3\)

b.

ĐKXD: \(x\ge-1\)

Phương trình: \(2\left(x+1\right)-\left(3x-2\right)\sqrt[]{x+1}+x^2-x=0\)

Đặt \(\sqrt[]{x+1}=t\ge0\)

\(\Rightarrow2t^2-\left(3x-2\right)t+x^2-x=0\)

\(\Delta=\left(3x-2\right)^2-8\left(x^2-x\right)=\left(x-2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x-2+x-2}{4}=x-1\\t=\dfrac{3x-2-x+2}{4}=\dfrac{x}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[]{x+1}=x-1\left(x\ge1\right)\\\sqrt[]{x+1}=\dfrac{x}{2}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=x^2-2x+1\left(x\ge1\right)\\x+1=\dfrac{x^2}{4}\left(x\ge0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2+2\sqrt[]{2}\end{matrix}\right.\)

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

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

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

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

<=> \(\orbr{\begin{cases}3.x+1=0\\2.x+10=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=-5\end{cases}}}\)

Vậy x = { \(\frac{-1}{3};-5\)} 

b) x² + 10.x + 25 - 4.x . ( x + 5 ) = 0 

<=> ( x + 5 )² -4.x . (x + 5 ) = 0

<=> ( x+ 5 ) . ( x + 5 - 4.x ) = 0

<=> ( x + 5 ) . ( 5 - 3.x )  = 0

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

Vậy x = \(\left\{\frac{5}{3};-5\right\}\)

c) (4.x - 5 )² - 2. ( 16.x² -25 ) = 0 

<=> ( 4.x-5)² -2 .( 4.x-5) .( 4.x + 5 ) = 0

<=> (  4.x -5 )² - ( 8.x+ 10 ) . ( 4.x -5 ) = 0

<=> ( 4.x -5 ) . ( 4.x-5 - 8.x - 10 ) = 0

<=> ( 4.x - 5 ) . ( -4.x - 15 ) = 0

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

Vậy x = \(\left\{\frac{5}{4};\frac{-15}{4}\right\}\)

d) ( 4.x + 3 )² = 4. ( x² - 2.x + 1 ) 

<=> 16.x² + 24.x + 9 - 4.x²  + 8.x - 4 = 0

<=> 12.x² + 32.x + 5 =0 

<=> 12. ( x +\(\frac{1}{8}\) ) . ( x + \(\frac{5}{2}\)) = 0 

<=> \(\orbr{\begin{cases}x+\frac{1}{6}=0\\x+\frac{5}{2}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{cases}}}\)

Vậy x = \(\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)

e) x² -11.x + 28 = 0

<=> x² -4.x  - 7.x + 28 = 0

<=> ( x - 7 ) . ( x - 4 ) = 0

<=> \(\orbr{\begin{cases}x-7=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=4\end{cases}}}\)

Vậy x = { 4 ; 7 } 

f ) 3.x.3 - 3.x² - 6.x = 0

<=> 3.x. ( x² -x - 2 ) = 0 

<=> 3.x. ( x - 2 ) . ( x + 1 ) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

        \([x=0\)                \([x=0\)

( Lưu ý :Lưu ý này không cần ghi vào vở :  Chị nối 2 ý đó làm 1 nha cj ! ) 

Vậy x = { 2 ; -1 ; 0 } 

a) 2x-7=11x+11

<=> 2x-11x=11+7

<=> -9x=17

<=> x= -17/9

b) 2011x -4 =x+6

<=> 2011x-x=6+4

<=> 2010x=10

<=> x=10/2010

<=> x=1/201

c) 5(2x-3)-2(3x-5)=0

<=> 10x-15-6x+10=0

<=> 10x-6x=15-10

<=>4x=5

<=> x=5/4