a có nè

cho mk nha

\(\left(\sqrt{x}-1+5\right)\left(x-6\sqrt{x}\right)\)

\(=\left(\sqrt{x}+4\right)\left(x-6\sqrt{x}\right)\)

\(=x\sqrt{x}-6x+4x-24\sqrt{x}\)

\(=x\sqrt{x}-2x-24\sqrt{x}\)

Đề bài?

Ta có:M=\(\frac{a^{10}b^7c^{2000}}{b^{2017}}\)=\(\frac{a^{10}}{b^{10}}\)x\(\frac{b^7}{b^7}\)x\(\frac{c^{2000}}{b^{2000}}\)=\(\left(\frac{a}{b}\right)^{10}\)x\(\left(\frac{c}{b}\right)^{2000}\)=\(\left(\frac{a}{b}\right)^{10}\)x\(\left(\frac{b}{c}\right)^{-2000}\)

Mà \(\frac{a}{b}\)=\(\frac{b}{c}\)nên M=\(\left(\frac{a}{b}\right)^{10}\)x\(\left(\frac{a}{b}\right)^{-2000}\)=\(\left(\frac{a}{b}\right)^{-1990}\)

 tinh m ma

\(\Rightarrow\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}=\frac{c}{\frac{1}{4}}\)

+ Áp dụng tính chất bằng nhau ta có : 

\(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}=\frac{c}{\frac{1}{4}}=\frac{a+b+c}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}=\frac{117}{\frac{13}{12}}=108\)

Suy ra \(\frac{a}{\frac{1}{2}}=108\Rightarrow a=54\)

              \(\frac{b}{\frac{1}{3}}=108\Rightarrow b=36\)

          \(\frac{c}{\frac{1}{4}}=108\Rightarrow c=27\)

Vậy \(a=54;b=36;c=27\)

Chúc bạn học tốt !!!

\(\Rightarrow a=2c;b=\frac{4}{3}c\Rightarrow a+b+c=\left(3+\frac{4}{3}\right)c=\frac{13}{3}=117\Rightarrow c=27\Rightarrow b=36;a=54\)

\(\frac{x}{3}=\frac{y}{5}=\frac{-4x}{-12}=\frac{3y}{15}=\frac{-4x+3y}{-12+15}=\frac{12}{3}=4\Rightarrow x=12;y=20\)

\(\Rightarrow\frac{-4x}{-12}=\frac{3y}{15}\)

+ Áp dụng tính chất bằng nhau ta có : 

\(\frac{-4x}{-12}=\frac{3y}{15}=\frac{-4x+3y}{-12+15}=\frac{12}{3}=4\)

Suy ra : \(\frac{-4x}{-12}=4\Rightarrow x=132\)

               \(\frac{3y}{15}=4\Rightarrow y=20\)

Vậy \(x=132;y=20\)

Chúc bạn học tốt !!!

Ta có: |2x - 5| \(\ge\)0 \(\forall\)x

=> |2x - 5| + 1,(3) \(\ge\)1,(3)

hay |2x - 5| + 4/3 \(\ge\)4/3

Dấu "=" xảy ra <=> 2x - 5 = 0 <=>  x = 5/2

Vậy Min F = 4/3 <=> x = 5/2

Ta có: G = |x - 3| + |x + 3/2|

G = |3 - x| + |x + 3/2| \(\ge\)|3 - x + x + 3/2| = |3/2| = 3/2

Dấu "=" xảy ra <=> (3 - x)(x + 3/2) \(\ge\)0

<=> -3/2 \(\le\)x \(\le\)3

Vậy MinG = 3/2 <=> -3/2 \(\le\)x \(\le\)3

Làm lại cho Edogawa Conan

\(G=\left|x-3\right|+\left|x+\frac{3}{2}\right|\)

\(G=\left|3-x\right|+\left|x+\frac{3}{2}\right|\ge\left|\left(3-x\right)+\left(x+\frac{3}{2}\right)\right|\)

\(=\frac{9}{2}\)

Vậy \(G_{min}=\frac{9}{2}\Leftrightarrow\left(3-x\right)\left(x+\frac{3}{2}\right)\ge0\)

\(Th1:\hept{\begin{cases}3-x\ge0\\x+\frac{3}{2}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge\frac{3}{2}\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le2\)

\(Th2:\hept{\begin{cases}3-x\le0\\x+\frac{3}{2}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le\frac{3}{2}\end{cases}}\left(L\right)\)

8\(\sqrt{x}\)= x^2

bình phương 2 vế, ta được:

64x = x^4

64x - x^4 = 0

x(64 - x³) = 0

x = 0 hoặc x = 4

\(8\sqrt{x}=x^2\)

\(\Leftrightarrow8\sqrt{x}-x^2=0\)

\(\Leftrightarrow\sqrt{x}\left(8-\sqrt{x^3}\right)=0\)

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