a) \(5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-11\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

Vậy ....

b) \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)

\(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)

\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\)

\(\Leftrightarrow2x^3-2x^2+8x=-16\)

\(\Leftrightarrow8x=-16\)

\(\Leftrightarrow x=-2\)

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

\(\Leftrightarrow2x^2-x-3=2x^2+9x-5\)

\(\Leftrightarrow10x-2=0\)

\(\Leftrightarrow x=\frac{1}{5}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{1}{5}\right\}\)

f) \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)

\(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)

\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\)

\(\Leftrightarrow3x+21=0\)

\(\Leftrightarrow x=-7\)

Vậy tập nghiệm của phương trình là \(S=\left\{-7\right\}\)

g) \(\left(x-1\right)-\left(2x-1\right)=9-x\)

\(\Leftrightarrow x-1-2x+1=9-x\)

\(\Leftrightarrow-x=9-x\)

\(\Leftrightarrow0=9\)(ktm)

Vậy tập nghiệm của phương trình là \(S=\varnothing\)

h) \(\left(x-3\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(\Leftrightarrow x^2+x-12-6x+4=x^2-8x+16\)

\(\Leftrightarrow x^2-5x-8=x^2-8x+16\)

\(\Leftrightarrow3x-24=0\)

\(\Leftrightarrow x=8\)

Vậy tập nghiệm của phương trình là \(S=\left\{8\right\}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow2abc\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Rightarrow2ab+2bc+2ac=0\)

Khi đó ta có:\(a+b+c=1\Rightarrow\left(a+b+c\right)^2=1\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=1\)

\(\Rightarrow a^2+b^2+c^2=1\left(ab+bc+ac=0\right)\left(đpcm\right)\)

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x = -100

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)

\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)

Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)

\(\Rightarrow200-x=0\Rightarrow x=200\)

Vậy x = 200

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Cs câu hỏi giống bn đó vào xem ik

Hoặc bn vào chỗ câu hỏi tương tự là dc nha

Học tốt !!!

Using correct form of words

1,____Charline Chaplin was born in London,England(legend)

2,In Canada,there are a lot of______from various countries(refuge)

3,Lake Wanaka is a_____scenery in New Zealand

:< rồi để căn nó mệt người mik đặt hem:P

Ta có: \(\hept{\begin{cases}\sqrt{a}=a\\\sqrt{b}=b\end{cases}}\)

\(P=a^2-2ab+3b^2-2a+1\)

\(\Leftrightarrow3P=3a^2-6ab+9b^2-6a+3\)

\(\Leftrightarrow3P=\left(x-3b\right)^2+2\left(a-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)

\(\Rightarrow P\ge\frac{1}{2}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a=\frac{3}{2}\\b=\frac{1}{2}\end{cases}}\) hay \(\hept{\begin{cases}a=\frac{9}{4}\\b=\frac{1}{4}\end{cases}}\)

Đặt \(\sqrt{a}=u;\sqrt{b}=v\left(u,v\ge0\right)\)

Lúc đó \(P=u^2-2uv+3v^2-2u+1\)

\(\Rightarrow3P=3u^2-6uv+9v^2-6u+3\)

\(=\left(u^2-6uv+9v^2\right)+2\left(u^2-6u+\frac{9}{4}\right)-\frac{3}{2}\)

\(=\left(u-3v\right)^2+2\left(u-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\)

\(\Rightarrow P\ge\frac{-1}{2}\)

(Dấu "=" khi \(\hept{\begin{cases}u=\frac{3}{2}\\v=\frac{1}{2}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\sqrt{a}=\frac{3}{2}\\\sqrt{b}=\frac{1}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{9}{4}\\b=\frac{1}{4}\end{cases}}\))

Ta có :

\(\left(x^2-2014\right)\left(x^2-2015\right)\left(x^2-2016\right)\)\(=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}x^2-2014=0\\x^2-2015=0\\x^2-2016=0\end{cases}}\)

Giải (1) :

    \(x^2-2014=0\)

     \(\hept{\begin{cases}x=\sqrt{2014}\\x=-\sqrt{2014}\end{cases}}\)

Giải (2) :

     \(x^2-2015=0\)

        \(\hept{\begin{cases}x=\sqrt{2015}\\x=-\sqrt{2015}\end{cases}}\)

Giải (3) :

   \(x^2-2016=0\)

    \(\hept{\begin{cases}x=\sqrt{2016}\\x=-\sqrt{2016}\end{cases}}\)

Vậy nghiệm nhỏ nhất của phương trình là \(x=-\sqrt{2016}\)

Chú ý : \(x^2-2014=0\)(1)

            \(x^2-2015=0\)(2)

            \(x^2-2016=0\)(3)

\(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}\)

\(=\frac{3x+1-2x+3}{x+y}\)

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

giúp mik vs

\(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}\)

\(=\frac{2\left(x^2-y^2\right)+2y^2}{x}\)

\(=\frac{2x^2-2y^2+2y^2}{x}\)

\(=\frac{2x^2}{x}\)

\(=2x\)