Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\sqrt{x}+\sqrt{\frac{x}{9}}-\frac{1}{3}\sqrt{4x}=5\)
ĐK : x ≥ 0
<=>\(\sqrt{x}+\sqrt{x\times\frac{1}{9}}-\frac{1}{3}\sqrt{2^2x}=5\)
<=> \(\sqrt{x}+\sqrt{x\times\left(\frac{1}{3}\right)^2}-\left(\frac{1}{3}\times\left|2\right|\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\left|\frac{1}{3}\right|\sqrt{x}-\left(\frac{1}{3}\times2\right)\sqrt{x}=5\)
<=> \(\sqrt{x}+\frac{1}{3}\sqrt{x}-\frac{2}{3}\sqrt{x}=5\)
<=> \(\sqrt{x}\left(1+\frac{1}{3}-\frac{2}{3}\right)=5\)
<=> \(\sqrt{x}\times\frac{2}{3}=5\)
<=> \(\sqrt{x}=\frac{15}{2}\)
<=> \(x=\frac{225}{4}\)( tm )
a) \(A=\frac{1}{2}\sqrt{32}+\sqrt{98}-\frac{1}{6}\sqrt{18}=\frac{1}{2}\sqrt{4^2.2}+\sqrt{7^2.2}-\frac{1}{6}.\sqrt{3^2.2}\)
\(=\frac{1}{2}\sqrt{4^2}.\sqrt{2}+\sqrt{7^2}.\sqrt{2}-\frac{1}{6}.\sqrt{3^2}.\sqrt{2}\)\(=\frac{1}{2}.4\sqrt{2}+7\sqrt{2}-\frac{1}{6}.3.\sqrt{2}\)\(=2.\sqrt{2}+7\sqrt{2}-\frac{1}{2}\sqrt{2}=\left(2+7-\frac{1}{2}\right)\sqrt{2}=\frac{17}{2}\sqrt{2}\)
a) Ta có \(\hept{\begin{cases}x^2\ge0\forall x\\\left(y-\frac{1}{3}\right)^2\ge0\forall y\end{cases}\Rightarrow}x^2+\left(y-\frac{1}{3}\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=0\\y-\frac{1}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=\frac{1}{3}\end{cases}}\)
Vậy x = 0 ; y = 1/3 là giá trị cần tìm
b) Ta có : \(\hept{\begin{cases}\left|2x-1\right|\ge0\forall x\\\left|x-3y+2\right|\ge0\forall x;y\end{cases}}\Rightarrow\left|2x-1\right|+\left|x-3y+2\right|\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-1=0\\x-3y+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\-3y=-\frac{3}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{2}\end{cases}}\)
Vạy \(x=y=\frac{1}{2}\)là giá trị cần tìm
a) Ta có : \(\hept{\begin{cases}x^2\ge0\forall x\\\left(y-\frac{1}{3}\right)^2\ge0\forall y\end{cases}}\Rightarrow x^2+\left(y-\frac{1}{3}\right)^2\ge0\forall x,y\)
Kết hợp với đề bài => Chỉ xảy ra trường hợp x2 + ( y - 1/3 )2 = 0
=> x = 0 ; y = 1/3
b) \(\hept{\begin{cases}\left|2x-1\right|\\\left|x-3y+2\right|\end{cases}\ge}0\forall x,y\Rightarrow\left|2x-1\right|+\left|x-3y+2\right|\ge0\forall x,y\)
Dấu "=" xảy ra khi x = 1/2 ; y = 5/6
a Ta có 4x2 - 4x + 3 = (4x2 - 4x + 1) + 2 = (2x - 1)2 + 2 \(\ge\)2 > 0 (đpcm)
b) Ta có y - y2 - 1
= -(y2 - y + 1)
= -(y2 - y + 1/4) - 3/4
= -(y - 1/2)2 - 3/4 \(\le-\frac{3}{4}< 0\)(đpcm)
a) 4x2 - 4x + 3 = ( 4x2 - 4x + 1 ) + 2 = ( 2x - 1 )2 + 2 ≥ 2 > 0 ∀ x ( đpcm )
b) y - y2 - 1 = -( y2 - y + 1/4 ) - 3/4 = -( y - 1/2 ) - 3/4 ≤ -3/4 < 0 ∀ x ( đpcm )
ĐK : 10x - 2020 \(\ge0\Rightarrow10x\ge2020\Rightarrow x\ge202\)
Với x \(\ge\)202
=> x - 1 > 0 ; x - 3 > 0 ; .... x - 35 > 0
Khi đó |x - 1| + |x - 3| + |x - 5| + ... + |x - 35| = 10x - 2020
= x - 1 + x - 3 + x - 5 + ... x - 35 = 10x - 2020
=> 18x - (1 + 3 + 5 + ... + 35) = 10x - 2020
=> 18x - 18(35 + 1) : 2 = 10x - 2020
=> 18x - 324 = 10x - 2020
=> 8x = -1696
=> x = -212 (loại)
Vạy \(x\in\varnothing\)
\(x^2+3x-10\)
\(=x^2+5x-2x-10\)
\(=\left(x^2+5x\right)-\left(2x+10\right)\)
\(=x\left(x+5\right)-2\left(x+5\right)\)
\(=\left(x-2\right)\left(x+5\right)\)
Thích hđt thì chiều :))
x2 + 3x - 10
= ( x2 + 3x + 9/4 ) - 49/4
= ( x + 3/2 )2 - ( 7/2 )2
= ( x + 3/2 - 7/2 )( x + 3/2 + 7/2 )
= ( x - 2 )( x + 5 )
\(\left(4x-8\right)^{11}=0\)
\(\Rightarrow4x-8=0\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=2\)
\(2^x+2^{x+3}=144\)
\(\Leftrightarrow2^x\left(1+2^3\right)=144\)
\(\Leftrightarrow2^x=16\)
\(\Leftrightarrow x=4\)
\(2^x+2^{x+3}=144\)
\(2^x+2^x.2^3=144\)
\(2^x.\left(1+2^3\right)=144\)
\(2^x.9=144\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
\(P=\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{4-6\sqrt{a}}{1-a}-\frac{-3}{\sqrt{a}+1}\)
ĐK : \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\)
a) \(P=\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{4-6\sqrt{a}}{a-1}+\frac{3}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}}{\sqrt{a}-1}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4-6\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}+4-6\sqrt{a}+3\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)
\(=\frac{a-2\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}+1}\)
Với \(a=4-2\sqrt{3}\)( tmđk \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
\(P=\frac{\sqrt{4-2\sqrt{3}}-1}{\sqrt{4-2\sqrt{3}}+1}\)
\(=\frac{\sqrt{3-2\sqrt{3}+1}-1}{\sqrt{3-2\sqrt{3}+1}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}-1}{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}+1^2}+1}\)
\(=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}-1}{\sqrt{\left(\sqrt{3}-1\right)^2}+1}\)
\(=\frac{\left|\sqrt{3}-1\right|-1}{\left|\sqrt{3}-1\right|+1}\)
\(=\frac{\sqrt{3}-1-1}{\sqrt{3}-1+1}=\frac{\sqrt{3}-2}{\sqrt{3}}\)
b) \(P=\frac{\sqrt{a}-1}{\sqrt{a}+1}=\frac{\sqrt{a}+1-2}{\sqrt{a}+1}=1-\frac{2}{\sqrt{a}+1}\)( ĐK \(\hept{\begin{cases}a\ge0\\a\ne1\end{cases}}\))
Để P đạt giá trị nguyên => \(\frac{2}{\sqrt{a}+1}\)nguyên
=> \(2⋮\sqrt{a}+1\)
=> \(\sqrt{a}+1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\sqrt{a}\in\left\{0;1\right\}\)< đã loại hai trường hợp âm >
=> \(a\in\left\{0\right\}\)< loại trường hợp a = 1 >
Vậy với a = 0 thì P có giá trị nguyên