x=? nếu 5x x = 10
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt{x^2-5x+14}=a\) và \(\sqrt{x^2-5x+10}=b\) \(\left(a,b>0\right)\)
\(\Rightarrow a-b=2\)
\(\Rightarrow a^2-b^2=x^2-5x+14-x^2+5x-10=4\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=4\)
\(\Leftrightarrow a-b=2\)
\(\Leftrightarrow\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}=2\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
\(5x{y^2}.C = 10{x^3}{y^3} \Rightarrow C = 10{x^3}{y^3}:5x{y^2} = 2{x^2}y\)
b)
\(\begin{array}{l}\left( {K + 5x{y^2}} \right).C = 6{x^4}y + 10{x^3}{y^3}\\ \Rightarrow \left( {K + 5x{y^2}} \right).2{x^2}y = 6{x^4}y + 10{x^3}{y^3}\\ \Rightarrow K + 5x{y^2} = \left( {6{x^4}y + 10{x^3}{y^3}} \right):2{x^2}y\\ \Rightarrow K + 5x{y^2} = 6{x^4}y:2{x^2}y + 10{x^3}{y^3}:2{x^2}y\\ \Rightarrow K + 5x{y^2} = 3{x^2} + 5x{y^2}\\ \Rightarrow K = 3{x^2} + 5x{y^2} - 5x{y^2}\\ \Rightarrow K = 3{x^2}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có
\(2A=\left(\sqrt{x^2-5x+14}-\sqrt{x^2-5x+10}\right)\left(\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}\right)\)
⇔ 2A=x2-5x+14-x2+5x-10
⇔2A= 4
⇔ A=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=x^2-4x+10=x^2-4x+4+6=\left(x-2\right)^2+6\ge6\)
Vậy GTNN A là 6 khi x - 2 = 0 <=> x = 2
\(B=\left(1-x\right)\left(3x-4\right)=3x-4-3x^2+4x=-3x^2+7x-4\)
\(=-3\left(x^2-\frac{7}{3}x+\frac{4}{3}\right)=-3\left(x^2-2.\frac{7}{6}x+\frac{49}{36}-\frac{1}{36}\right)=-3\left(x-\frac{7}{6}\right)^2+\frac{1}{12}\ge\frac{1}{12}\)
\(=3\left(x-\frac{7}{6}\right)^2-\frac{1}{12}\le-\frac{1}{12}\)Vậy GTLN B là -1/12 khi x = 7/6
\(C=3x^2-9x+5=3\left(x^2-3x+\frac{5}{3}\right)=3\left(x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{7}{12}\right)\)
\(=3\left(x-\frac{3}{2}\right)^2-\frac{7}{4}\ge-\frac{7}{4}\)Vậy GTNN C là -7/4 khi x = 3/2
\(D=-2x^2+5x+2=-2\left(x^2-\frac{5}{2}x-1\right)=-2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}-\frac{41}{16}\right)\)
\(=-2\left(x-\frac{5}{4}\right)^2+\frac{21}{8}\le\frac{21}{8}\)Vậy GTLN D là 21/8 khi x = 5/4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) | x + 10 | - ( 5 - 3x ) = ( 4x - 10 ) - ( x - 5 )
=> | x + 10 | = ( 5 - 3x ) + ( 4x - 10 ) - ( x - 5 )
=> | x + 10 | = 5 - 3x + 4x - 10 - x + 5
=> | x +10 | = 0
=> x + 10 = 0
=> x = -10
Vậy...
b) Làm tương tự
Kết quả : | 3x + 21 | = -10 ( vô lí) ( vì |3x+21| >= 0 mà -10<0)
Vậy không tìm được x thỏa mãn bài toán
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-10\right)^2-2\left(x-10\right)\left(5x-3\right)+\left(5x-3\right)^2\)
\(=\left(x-10-5x+3\right)^2=\left(-4x-7\right)^2=\left(4x+7\right)^2\)
\(=16x^2+56x+49\)
Ta có: \(\left(x-10\right)^2-2\left(x-10\right)\left(5x-3\right)+\left(5x-3\right)^2\)
\(=\left(x-10-5x+3\right)^2\)
\(=\left(-4x+6\right)^2\)
\(=16x^2-48x+36\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
(\(\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}\))(\(\sqrt{x^2-5x+14}-\sqrt{x^2-5x+10}\)) = 4
=> M = 2
X BANG 2
5 * x= 10
x= 10:5
x = 2
tk mk nha mk tk lại cho