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Bài 1:
a)
\(|x+\frac{4}{15}|-|-3,75|=-|-2,15|\)
\(\Leftrightarrow |x+\frac{4}{15}|-3,75=-2,15\)
\(\Leftrightarrow |x+\frac{4}{15}|=-2,15+3,75=\frac{8}{5}\)
\(\Rightarrow \left[\begin{matrix} x+\frac{4}{15}=\frac{8}{5}\\ x+\frac{4}{15}=-\frac{8}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{4}{3}\\ x=\frac{-28}{15}\end{matrix}\right.\)
b )
\(|\frac{5}{3}x|=|-\frac{1}{6}|=\frac{1}{6}\)
\(\Rightarrow \left[\begin{matrix} \frac{5}{3}x=\frac{1}{6}\\ \frac{5}{3}x=-\frac{1}{6}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{10}\\ x=-\frac{1}{10}\end{matrix}\right.\)
c)
\(|\frac{3}{4}x-\frac{3}{4}|-\frac{3}{4}=|-\frac{3}{4}|=\frac{3}{4}\)
\(\Leftrightarrow |\frac{3}{4}x-\frac{3}{4}|=\frac{3}{2}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}x-\frac{3}{4}=\frac{3}{2}\\ \frac{3}{4}x-\frac{3}{4}=-\frac{3}{2}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=-1\end{matrix}\right.\)
Bài 3:
a) Ta thấy:
\(|x+\frac{15}{19}|\geq 0, \forall x\Rightarrow A\ge 0-1=-1\)
Vậy GTNN của $A$ là $-1$ khi \(x+\frac{15}{19}=0\Leftrightarrow x=-\frac{15}{19}\)
b)Vì \(|x-\frac{4}{7}|\geq 0, \forall x\Rightarrow B\geq \frac{1}{2}+0=\frac{1}{2}\)
Vậy GTNN của $B$ là $\frac{1}{2}$ khi \(x-\frac{4}{7}=0\Leftrightarrow x=\frac{4}{7}\)
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Bài 1:
a) \(\left|x-\dfrac{2}{3}\right|+\left|y+x\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left|x-\dfrac{2}{3}\right|=0\\\left|y+x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\y+x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{2}{3}\end{matrix}\right.\)
b) \(\left(x-2y\right)^2+\left|x+\dfrac{1}{6}\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)^2=0\\\left|x+\dfrac{1}{6}\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2y=0\\x+\dfrac{1}{6}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=x\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y=-\dfrac{1}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{1}{12}\\x=\dfrac{-1}{6}\end{matrix}\right.\)
a: \(B=\left|2-x\right|+1.5>=1.5\)
Dấu '=' xảy ra khi x=2
b: \(B=-5\left|1-4x\right|-1\le-1\)
Dấu '=' xảy ra khi x=1/4
g: \(C=x^2+\left|y-2\right|-5>=-5\)
Dấu '=' xảy ra khi x=0 và y=2
1. Ta có: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}=x+y+z\) ( vì \(a+b+c=1\) )
Do đó \(\left(x+y+z\right)^2=\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\)( vì \(a^2+b^2+c^2=1\) ).
Vậy \(\left(x+y+z\right)^2=x^2+y^2+z^2\)
2. Đặt \(x^2=a\left(a\ge0\right),y^2=b\left(b\ge0\right)\)
Ta có: \(\dfrac{a+b}{10}=\dfrac{a-2b}{7}\) và \(a^2b^2=81\)
\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}=\dfrac{\left(a+b\right)-\left(a-2b\right)}{10-7}=\dfrac{3b}{3}=b\) __(1)__
\(\dfrac{a+b}{10}=\dfrac{a-2b}{7}=\dfrac{2a+2b}{20}=\dfrac{\left(2a+2b\right)+\left(a-2b\right)}{20+7}=\dfrac{3a}{27}=\dfrac{a}{9}\)__(2)__
Từ (1) và (2) suy ra \(\dfrac{a}{9}=b\Rightarrow a=9b\)
Do \(a^2b^2=81\) nên \(\left(9b\right)^2.b^2=81\Rightarrow81b^4=81\Rightarrow b^4=1\Rightarrow b=1\) ( vì \(b\ge0\) )
Suy ra: a = 9.1 = 9
Ta có: \(x^2=9\) và \(y^2=1\). Suy ra: \(x=\pm3,y=\pm1\)
1. Tìm x:
a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80
b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)
c) Xem lại đề
d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)
e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2
3. Tính giá trị của biểu thức:
\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)
\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)
\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)
a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
Bài 1:
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|x+y\right|\ge0\forall x,y\end{matrix}\right.\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|x+y\right|\ge0\forall x,y\)
Vì vậy, để tìm được x, y thỏa mãn đề bài thì \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=0\\x+y=0\end{matrix}\right.\)
Từ đó, ta tìm được \(x=\dfrac{1}{2}\) và \(y=-\dfrac{1}{2}\)
Bài 2:
\(A=\left|x-\dfrac{3}{4}\right|\)
Ta thấy \(\left|x-\dfrac{3}{4}\right|\ge0\forall x\Rightarrow A\ge0\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=0\Leftrightarrow x-\dfrac{3}{4}=0\Leftrightarrow x=\dfrac{3}{4}\)
Vậy GTNN của A là 0 khi \(x=\dfrac{3}{4}\)
\(B=\left|x+\dfrac{2}{3}\right|+2\)
\(\left|x+\dfrac{2}{3}\right|\ge0\forall x\) nên \(\left|x+\dfrac{2}{3}\right|+2\ge2\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left|x+\dfrac{2}{3}\right|=0\Leftrightarrow x+\dfrac{2}{3}=0\Leftrightarrow x=-\dfrac{2}{3}\)
Vậy GTNN của B là 2 khi \(x=-\dfrac{2}{3}\)