Giai các phương trình
a, |x-3| + |5-x|=2a
b) 2|x+a|-|x-2a|=3a
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a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
https://hoc24.vn/hoi-dap/question/177629.html (câu 2)
https://hoc24.vn/hoi-dap/question/940816.html?pos=2486212 (câu 1)a. \(x^2-2\sqrt{5}x+5=0\)
<=> \(x^2-2x\sqrt{5}+\left(\sqrt{5}\right)^2=0\)
<=> \(\left(x-\sqrt{5}\right)^2=0\)
<=> \(x-\sqrt{5}=0\)
<=> \(x=\sqrt{5}\)
b. \(\sqrt{x+3}=1\) ĐK: x \(\ge-3\)
<=> x + 3 = 12
<=> x = 1 - 3
<=> x = -2 (TM)
a: Ta có: \(x^2-2x\sqrt{5}+5=0\)
\(\Leftrightarrow x-\sqrt{5}=0\)
hay \(x=\sqrt{5}\)
b: Ta có: \(\sqrt{x+3}=1\)
\(\Leftrightarrow x+3=1\)
hay x=-2
Ta có: \(\hept{\begin{cases}xy+x+y=1\\yz+y+z=3\\xz+x+z=7\end{cases}}\Rightarrow\hept{\begin{cases}xy+x+y+1=2\\yz+y+z+1=4\\xz+x+z+1=8\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+1\right)\left(y+1\right)=2\\\left(y+1\right)\left(z+1\right)=4\\\left(x+z\right)\left(z+1\right)=8\end{cases}}\)
Nhân theo vế:
\(\left[\left(x+1\right)\left(y+1\right)\left(z+1\right)\right]^2=64\Rightarrow\orbr{\begin{cases}\left(x+1\right)\left(y+1\right)\left(z+1\right)=8\\\left(x+1\right)\left(y+1\right)\left(z+1\right)=-8\end{cases}}\)
Thay vào từng trường hợp tìm x;y;z
Lời giải:
Vì $|x+4|, |x+5|\geq 0$ với mọi $x\in\mathbb{R}$ nên:
$2x=|x+4|+|x+5|\geq 0$
$\Rightarrow x\geq 0$
$\Rightarrow |x+4|=x+4; |x+5|=x+5$. Do đó, pt trở thành:
$x+4+x+5=2x$
$\Leftrightarrow 0=9$ (vô lý)
Vậy pt vô nghiệm.
b)
Ta có:
$2x=|x+4|+|x+5|+...+|x+10|\geq 0$
$\Rightarrow x\geq 0$
$\Rightarrow |x+4|=x+4; |x+5|=x+5; ....;|x+10|=x+10$
Do đó pt trở thành:
$2x=(x+4)+(x+5)+...+(x+10)$
$2x=7x+49$
$x=\frac{-49}{5}<0$ (vô lý vì $x\geq 0$)
Vậy PT vô nghiệm.
\(\text{2x - (x - 3)(5 - x) = (x+4)}^2.\)
\(\Leftrightarrow2x-\left(5x-x^2-15+3x\right)=x^2+8x+16.\)
\(\Leftrightarrow2x-5x+x^2+15-3x-x^2-8x-16=0.\)
\(\Leftrightarrow-14x-1=0.\Leftrightarrow x=\dfrac{-1}{14}.\)
\(\text{(4x + 1)(x - 2) + 25 = (2x+3)}^2-4x.\)
\(\Leftrightarrow4x^2-8x+x-2+25=4x^2+12x+9-4x.\)
\(\Leftrightarrow-15x+14=0.\Leftrightarrow x=\dfrac{14}{15}.\)
xin loi
minh chang hieu cai gi ca
nho k minh nha
minh k lai cho
minh chang hieu gi ca
a=28
b=36