càng cấm thì càng phải làm :))

a,\(5x+2-x=22\)

\(< =>4x=22-2=20\)

\(< =>x=\frac{20}{4}=\frac{10}{2}=5\)

b, \(-5x+x-5=-13\)

\(< =>-4x=-13+5=-8\)

\(< =>x=\frac{-8}{-4}=\frac{4}{2}=2\)

c, \(\hept{\begin{cases}5x-6y=-30\left(+\right)\\x+y+x^2-y=42\left(++\right)\end{cases}}\)

Ta có : \(\left(++\right)\Leftrightarrow x^2+x=42< =>x^2+x-42=0\)

\(< =>x^2+7x-6x-42=0\)

\(< =>x\left(x+7\right)-6\left(x+7\right)=0\)

\(< =>\left(x-6\right)\left(x+7\right)=0< =>\orbr{\begin{cases}x=6\\x=-7\end{cases}}\)

Với \(x=6\)thì \(\left(+\right)\Leftrightarrow30-6y=-30< =>6y=60\)

\(< =>y=\frac{60}{6}=10\)

Với \(x=-7\)thì \(\left(+\right)\Leftrightarrow-35-6y=-30< =>6y=5< = >y=\frac{5}{6}\)

Vậy ta có hai cặp số x,y thỏa mãn hệ phương trình sau : \(\left\{6;10\right\};\left\{-7;\frac{5}{6}\right\}\)

a, \(5x+2-x=22\Leftrightarrow4x-20=0\Leftrightarrow x=5\)

b, \(-5x+x-5=-13\Leftrightarrow-4x+8=0\Leftrightarrow x=2\)

c, \(\hept{\begin{cases}5x-6y=-30\\x+y+x^2-y=42\end{cases}\Leftrightarrow\hept{\begin{cases}5x-6y=-30\\x+x^2-42=0\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}5x-6y=-30\\\left(x-6\right)\left(x+7\right)=0\end{cases}\Leftrightarrow\hept{\begin{cases}5x-6y=-30\left(k\right)\\x=6;-7\end{cases}}}\)

Thay vào pt k với x = 6 : \(5.6-6y=-30\Leftrightarrow30-6y+30=0\Leftrightarrow y=10\)

Thay vào pt k với x = -7 : \(5.\left(-7\right)-6y=-30\Leftrightarrow-35-6y=-30\Leftrightarrow y=\frac{5}{6}\)

Ta có: 

\(3x^4+1=x^4+x^4+x^4+1\ge4\sqrt[4]{x^4.x^4.x^4.1}=4x^3\)

Tương tự: \(3y^4+1\ge4y^3\) ; \(3z^4+1\ge4z^3\)

=> \(3\left(x^4+y^4+z^4\right)+3\ge4\left(x^3+y^3+z^3\right)\) (1)

Thay vào:

\(A=x^2\left(x+y\right)+y^2\left(y+z\right)+z^2\left(z+x\right)\)

\(A=x^3+x^2y+y^3+y^2z+z^3+z^2x\)

\(A=x^3+y^3+z^3+\left(x^2y+y^2z+z^2x\right)\)

\(\le x^3+y^3+z^3+\left(\frac{x^3+x^3+y^3+y^3+y^3+z^3+z^3+z^3+x^3}{3}\right)\)

\(=2\left(x^3+y^3+z^3\right)\)

\(=\frac{1}{2}\left[4\left(x^3+y^3+z^3\right)\right]\le\frac{1}{2}\left[3\left(x^4+y^4+z^4\right)+3\right]\)

\(=\frac{1}{2}\left[3.3+3\right]=\frac{12}{2}=6\)

Dấu "=" xảy ra khi: \(x=y=z=1\)

Vậy Max(A) = 6 khi x = y = z = 1

a) \(\sqrt{1-x}=\sqrt[3]{8}\) ( ĐK: \(x\le1\) )

\(\Leftrightarrow\sqrt{1-x}=2\)

\(\Leftrightarrow1-x=4\)

\(\Leftrightarrow x=-3\) ( Thỏa mãn )

b) \(\sqrt{4x^2-12x+9}=x+1\) ( ĐK : \(x\ge-1\) )

\(\Leftrightarrow\sqrt{\left(2x\right)^2-2.2x.3+3^2}=x+1\)

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

\(\Leftrightarrow\left|2x-3\right|=x+1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\\3-2x=x+1\end{cases}}\) \(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}}\) ( Thỏa mãn )

c) \(x+\sqrt{x}-2=0\) ( ĐK : \(x\ge0\) )

\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\sqrt{x}-1=0\)

\(\Leftrightarrow x=1\) ( Thỏa mãn )

+) ĐKXĐ : \(x\le1\)

 \(\sqrt{1-x}=\sqrt[3]{8}\)

\(\Leftrightarrow\sqrt{1-x}=2\)

\(\Leftrightarrow1-x=4\)

\(\Leftrightarrow x=-3\left(TM\right)\)

+)  \(\sqrt{4x^2-12x+9}=x+1\)

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

\(\Leftrightarrow\left|2x-3\right|=x+1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=x+1\left(x\ge\frac{3}{2}\right)\\2x-3=-x-1\left(x< \frac{3}{2}\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x=3+1\\2x+x=3-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\3x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{2}{3}\end{cases}\left(TM\right)}}\)

+) ĐKXĐ : \(x\ge0\)

 \(x+\sqrt{x}-2=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=2\)

+) \(\hept{\begin{cases}\sqrt{x}=1\\\sqrt{x}+1=2\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=1\end{cases}\Leftrightarrow}x=1\left(TM\right)}\)

+) \(\hept{\begin{cases}\sqrt{x}=2\\\sqrt{x}+1=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{2}\\x=0\end{cases}}}\left(TM\right)\)

Bài làm:

1) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)

\(=3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)

\(=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)

\(=14\sqrt{3}\)

2) \(\left(\sqrt{45}-2\sqrt{10}+\sqrt{5}\right).\sqrt{5}+5\sqrt{8}\)

\(=3\sqrt{5}.\sqrt{5}-2\sqrt{10}.\sqrt{5}+\sqrt{5}.\sqrt{5}+5.2\sqrt{2}\)

\(=15-10\sqrt{2}+5+10\sqrt{2}\)

\(=20\)

\(Q=\frac{1+\text{ax}}{1-\text{ax}}\sqrt{\frac{1-bx}{1+bx}}\)

Ta có: \(x=\frac{1}{a}\sqrt{\frac{2a-b}{b}}\Rightarrow\text{ax}=\sqrt{\frac{2a-b}{b}}\Rightarrow1+\text{ax}=1+\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}}\)

\(1-\text{ax}=\frac{\sqrt{b}-\sqrt{2a-b}}{\sqrt{b}}\)

\(\Rightarrow\frac{1+\text{ax}}{1-\text{ax}}=\frac{\sqrt{b}+\sqrt{2a-b}}{\sqrt{b}-\sqrt{2a-b}}=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2b-2a}\left(1\right)\)

 \(bx=\frac{b}{a}\sqrt{\frac{2a-b}{b}}=\frac{\sqrt{b}\left(2a-b\right)}{a}\Rightarrow\hept{\begin{cases}1-bx=\frac{a-\sqrt{b\left(2a-b\right)}}{a}\\1+bx=\frac{a+\sqrt{b\left(2a-b\right)}}{a}\end{cases}}\)

\(\Rightarrow\frac{1-bx}{1+bx}=\frac{a-\sqrt{b\left(2a-b\right)}}{a+\sqrt{b\left(2a-b\right)}}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{a^2-2ab+b^2}=\frac{\left(a-\sqrt{b\left(2a-b\right)}\right)^2}{\left(a-b\right)^2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow Q=\frac{\left(\sqrt{b}+\sqrt{2a-b}\right)^2}{2\left(b-a\right)}.\frac{a-\sqrt{b\left(2a-b\right)}}{a-b}=\frac{\text{[}2a+2\sqrt{b\left(2a-b\right)}\text{]}\left(a-b\sqrt{2a-b}\right)}{2\left(a-b\right)^2}\)

\(\Rightarrow\frac{2\left[a^2-b\left(2a-b\right)\right]}{2\left(a-b\right)^2}=\frac{2\left(a^2-2ab+b^2\right)}{a\left(a-b\right)^2}=1\)

ta có \(A=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}>0\forall x>0;x\ne1\left(1\right)\)

\(A-2=\frac{\sqrt{x}+2}{x+\sqrt{x}+1}-2=\frac{\sqrt{x}+2-2\sqrt{x}-2x-2}{\sqrt{x}+x+1}=\frac{-\sqrt{x}-2x}{\sqrt{x}+x+1}\le0\forall x\)

=> A =< 2 (2)

Từ (1) và (2) => 0<A =< 2 => A={1;2}

với A=1 \(\Rightarrow\frac{\sqrt{x}+2}{x+\sqrt{x}+1}=1\Leftrightarrow\sqrt{x}+2=x+\sqrt{x}+1\Leftrightarrow x=1\left(ktm\right)\)

với A=2\(\Rightarrow\frac{\sqrt{x}+2}{x+\sqrt{x}+1}=2\Leftrightarrow\sqrt{x}+2=2\sqrt{x}+2x+2\)

\(\Leftrightarrow2x+\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)=0\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(ktm\right)\)

Vậy không có giá trị x để A là số nguyên

\(\left(\sqrt{6}-2\sqrt{3}+5\sqrt{2}-\frac{1}{2\sqrt{8}}\right).2\sqrt{6}\)

\(=2.6-12\sqrt{3}+20\sqrt{3}-\frac{\sqrt{3}}{2}\)

\(=\frac{24+15\sqrt{3}}{2}\)

\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}=6\sqrt{5}-6\sqrt{5}+4\sqrt{5}=4\sqrt{5}\)

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}=\left(2\sqrt{7}-2\sqrt{2}.\sqrt{7}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+14\sqrt{2}=21\)

\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}=6\sqrt{3}-12\sqrt{3}+20\sqrt{3}=14\sqrt{3}\)

câu tiếp tương tự câu thứ 2 nha

Cho hàm số bậc nhất y = ax + 7

Thế x = 7 , y = 8 vào hàm số ta được :

8 = a.7 + 7

<=> 8 = 7a + 7

<=> 7a = 1

<=> a = 1/7

Vậy a = 1/7 

cho hàm số bậc nhất y= ax + 7 

thế x=7, y=8 vào hàm số ta được :

<=> 8=7a + 7

<=> 7a=1

<=> a=1/7 

vậy a=1/7   @_@ cảm ơn bạn