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đề sai ko vậy bạn
nếu đề đúng thì mình nghỉ là
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+3+8\sqrt{2x-13}}=5\)
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+3+8\sqrt{2x-13}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-13+8\sqrt{2x-13}+16}=5\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-13}+4\right)^2}=5\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-13}+4=5\)
\(\Leftrightarrow\sqrt{2x-3}+\sqrt{2x-13}=0\left(vl\right)\)
suy ra pt vô nghiệm
theo tôi là vậy
Ta có: \(\sqrt{2x-2+2\sqrt{2x-3}+\sqrt{2x+13+8\sqrt{2x-3}}}=5\)
\(\Leftrightarrow\sqrt{2x-2+2\sqrt{2x-3}+2\sqrt{2x-3}+4}=5\)
\(\Leftrightarrow\sqrt{2x+2+4\sqrt{2x-3}}=5\)
\(\Leftrightarrow\sqrt{2x-3+2\cdot\sqrt{2x-3}\cdot2+4+1}=5\)
\(\Leftrightarrow\left(\sqrt{2x-3}+2\right)^2+1=25\)
\(\Leftrightarrow\left(\sqrt{2x-3}+2\right)^2=24\)
\(\Leftrightarrow\sqrt{2x-3}+2=2\sqrt{6}\)
\(\Leftrightarrow2x-3=\left(2\sqrt{6}-2\right)^2\)
\(\Leftrightarrow2x-3=28-8\sqrt{6}\)
\(\Leftrightarrow2x=31-8\sqrt{6}\)
hay \(x=\dfrac{31-8\sqrt{6}}{2}\)
`\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8sqrt{2x-3}}=5(x>=3/2)`
`<=>\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=5`
`<=>\sqrt{(\sqrt{2x-3}+1)^2}+\sqrt{(\sqrt{2x-3}+4)^2}=5`
`<=>\sqrt{2x-3}+1+\sqrt{2x-3}+4=5`
`<=>2\sqrt{2x-3}=0`
`<=>\sqrt{2x-3}=0<=>2x-3=0<=>x=3/2(tmdk)`
Vậy `S={3/2}`
\(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13-8\sqrt{2x-3}}=5\\ \Leftrightarrow\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3-8\sqrt{2x-3}+16}=5\\ \Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}-4\right)^2}=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}-4\right|=5\\ \Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|=5\)
Có \(\left|\sqrt{2x-3}+1\right|+\left|4-\sqrt{2x-3}\right|\ge\left|\sqrt{2x-3}+1+4-\sqrt{2x-3}\right|=\left|5\right|=5\)
Dấu "=" xảy ra ⇔ Đẳng thức ban đầu xảy ra \(\Leftrightarrow\left(\sqrt{2x-3}+1\right)\left(4-\sqrt{2x-3}\right)=0\\ \Leftrightarrow4\sqrt{2x-3}-2x+3+4-\sqrt{2x-3}=0\\ \Leftrightarrow3\sqrt{2x-3}=2x-7\\ \Leftrightarrow\sqrt{2x-3}=\dfrac{2x-7}{3}\left(ĐK:x\ge\dfrac{7}{2}\right)\\ \Leftrightarrow2x-3=\dfrac{\left(2x-7\right)^2}{9}\\ \Leftrightarrow\left(2x-7\right)^2=9\left(2x-3\right)\\ \Leftrightarrow4x^2-28x+49-18x+27=0\\ \Leftrightarrow4x^2-40x+76=0\\ \Leftrightarrow x^2-10x+19=0\\ \Leftrightarrow\left(x^2-10x+25\right)-6=0\\ \Leftrightarrow\left(x-5\right)^2-\left(\sqrt{6}\right)^2=0\\ \Leftrightarrow\left(x-5-\sqrt{6}\right)\left(x-5+\sqrt{6}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{6}\left(tmđk\right)\\x=5-\sqrt{6}\left(ktmđk\right)\end{matrix}\right.\)
Vậy \(x=5+\sqrt{6}\) là nghiệm của pt.
a: 2x-3/2+3/4=-4
=>2x-3/4=-4
=>2x=-13/4
hay x=-13/8
b: \(\left(-\dfrac{2}{3}x-\dfrac{3}{5}\right)\cdot\left(\dfrac{-3}{2}-\dfrac{10}{3}\right)=\dfrac{2}{5}\)
\(\Leftrightarrow-\dfrac{2}{3}x-\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{-29}{6}=\dfrac{-2}{5}\cdot\dfrac{6}{29}=\dfrac{-12}{145}\)
=>2/3x+3/5=12/145
=>2/3x=-15/29
hay x=-45/58
c: \(\dfrac{x}{2}-\left(\dfrac{3}{5}x-\dfrac{13}{5}\right)=-\left(\dfrac{7}{10}x+\dfrac{7}{5}\right)\)
=>1/2x-3/5x+13/5=-7/10x-7/5
=>-1/10x+7/10x=-7/5-13/5
=>3/5x=-2
hay x=-2:3/5=-10/3
b. `|x + 1| + |2x - 3| = |3x - 2|`
Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)
\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)
Vậy phương trình có vô số nghiệm.
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2