Ta có : \(\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+2=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-2\end{array}\right.\)

Vậy \(x\in\left\{1;-2\right\}\)

Đây giống bài lớp 6 hơn

(x-1)(x+2)=0

=>x-1=0 hoặc x+2=0

=>x=1 hoặc x=-2

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

\(\Rightarrow x-1=0\) hoặc \(x+2=0\)

+) \(x-1=0\Rightarrow x=1\)

+) \(x+2=0\Rightarrow x=-2\)

Vậy x = 1 hoặc x = -2

x=1 hoặc x=-1

x=-1 hoặc x=1

cách làm sao bạn

2006 . | x - 1 | + ( x - 1 )² = 2005 . | 1 - x |

\(\Rightarrow\)2006 . | x - 1 | + ( x - 1 )² - 2005 . | 1 - x | = 0

Mà | x - 1 | = | 1 - x | = x - 1 

Thay vào , ta được :

2006 . ( x - 1 ) + ( x - 1 )² - 2005 . ( x - 1 ) = 0

( 2006 - 2005 ) . (x  - 1 ) + ( x - 1 )² = 0

( x - 1 ) + ( x - 1 )² = 0

vì ( x - 1 )² \(\ge\)0 

\(\Rightarrow\hept{\begin{cases}x-1=0\\\left(x-1\right)^2=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\left(tm\right)}\)

Vậy x = 1

\(\frac{\left(x-1\right)\left(x+5\right)}{\left(x-1\right)\left(2x+6\right)}=1\) <=> (x - 1)(x + 5) = (x - 1)(2x + 6)

<=> x + 5 = 2x + 6 (cùng chia cả 2 vế cho x - 1)

<=> 0 = x + 1 (cùng bớt cả 2 vế đi x và 5)

<=. x = 0 - 1

<=> x = -1

Ta có: \(\frac{1}{\left(3x+1\right)\left(y+z\right)+x}=\frac{1}{3x\left(y+z\right)+x+y+z}\le\frac{1}{3x\left(y+z\right)+3\sqrt[3]{xyz}}\)

\(=\frac{1}{3x\left(y+z\right)+3\sqrt[3]{1}}=\frac{1}{3x\left(y+z\right)+3}=\frac{1}{3\left(xy+zx+1\right)}=\frac{1}{3}\cdot\frac{1}{\frac{1}{y}+\frac{1}{z}+1}\)

Tương tự ta chứng minh được:

\(\frac{1}{\left(3y+1\right)\left(z+x\right)+y}\le\frac{1}{3}\cdot\frac{1}{\frac{1}{z}+\frac{1}{x}+1}\) ; \(\frac{1}{\left(3z+1\right)\left(x+y\right)+z}\le\frac{1}{3}\cdot\frac{1}{\frac{1}{x}+\frac{1}{y}+1}\)

Cộng vế 3 BĐT trên lại:

\(A\le\frac{1}{3}\cdot\left(\frac{1}{\frac{1}{x}+\frac{1}{y}+1}+\frac{1}{\frac{1}{y}+\frac{1}{z}+1}+\frac{1}{\frac{1}{z}+\frac{1}{x}+1}\right)\)

\(\Leftrightarrow3A\le\frac{1}{\left(\frac{1}{\sqrt[3]{x}}\right)^3+\left(\frac{1}{\sqrt[3]{y}}\right)^3+1}+\frac{1}{\left(\frac{1}{\sqrt[3]{y}}\right)^3+\left(\frac{1}{\sqrt[3]{z}}\right)^3+1}+\frac{1}{\left(\frac{1}{\sqrt[3]{z}}\right)^3+\left(\frac{1}{\sqrt[3]{x}}\right)^3+1}\)

Đặt \(\left(\frac{1}{\sqrt[3]{x}};\frac{1}{\sqrt[3]{y}};\frac{1}{\sqrt[3]{z}}\right)=\left(a;b;c\right)\) khi đó:

\(3A\le\frac{1}{a^3+b^3+1}+\frac{1}{b^3+c^3+1}+\frac{1}{c^3+a^3+1}\)

\(=\frac{1}{\left(a+b\right)\left(a^2-ab+b^2\right)+1}+\frac{1}{\left(b+c\right)\left(b^2-bc+c^2\right)+1}+\frac{1}{\left(c+a\right)\left(c^2-ca+a^2\right)+1}\)

\(\le\frac{1}{\left(a+b\right)\left(2ab-ab\right)+1}+\frac{1}{\left(b+c\right)\left(2bc-bc\right)+1}+\frac{1}{\left(c+a\right)\left(2ca-ca\right)+1}\)

\(=\frac{1}{ab\left(a+b\right)+1}+\frac{1}{bc\left(b+c\right)+1}+\frac{1}{ca\left(c+a\right)+1}\)

\(=\frac{abc}{ab\left(a+b\right)+abc}+\frac{abc}{bc\left(b+c\right)+abc}+\frac{abc}{ca\left(c+a\right)+abc}\)

\(=\frac{c}{a+b+c}+\frac{a}{b+c+a}+\frac{b}{c+a+b}\)

\(=\frac{a+b+c}{a+b+c}=1\)

Dấu "=" xảy ra khi: \(a=b=c\Leftrightarrow x=y=z=1\)

Vậy Max(A) = 1 khi x = y = z = 1

Câu hỏi của Pham Van Hung - Toán lớp 9 - Học toán với OnlineMath

a) Ta có : \(1+x^2=xy+yz+zx+x^2=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(z+x\right)\)

b) \(\Sigma\left(x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\right)=\Sigma\left(x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right).\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\right)\)

\(=\Sigma\left(x\left(y+z\right)\right)=xy+xz+xy+yz+zx+zy=2\left(xy+yz+zx\right)=2\)