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\(\left(x+3\right)^{2022}+\left(\sqrt{y-2}-1\right)^{2023}=0\) \(\left(ĐKXĐ: y\ge2\right)\)
Xét \(\left(x+3\right)^{2022}\ge0\forall x\)
\(\Rightarrow\left(\sqrt{y-2}-1\right)^{2023}\le0\)
\(\Leftrightarrow\sqrt{y-2}-1\le0\)
\(\Leftrightarrow\sqrt{y-2}\le1\)
\(\Leftrightarrow y-2\le1\)
\(\Rightarrow y\le3\)
\(\Rightarrow2\le y\le3\) mà \(y\in Z\)
\(\Rightarrow\left\{{}\begin{matrix}y=2\Leftrightarrow x=-2\\y=3\Leftrightarrow x=-3\end{matrix}\right.\)
Em không nghĩ câu này đúng. Anh giải thích hộ bạn đó với ạ.
1.
a) \(\frac{11}{2}-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=3-\frac{11}{2}\)
\(-\frac{2}{3}:\left|2x+-\frac{3}{2}\right|=-\frac{5}{2}\)
\(\left|2x+-\frac{3}{2}\right|=-\frac{2}{3}:\left(-\frac{5}{2}\right)\)
\(\left|2x+-\frac{3}{2}\right|=\frac{4}{15}\)
\(\Rightarrow\left|2x+-\frac{3}{2}\right|\in\text{{}\frac{4}{15};-\frac{4}{15}\)}
Nếu, \(2x+\left(-\frac{3}{2}\right)=\frac{4}{15}\)
\(2x=\frac{53}{30}\)
\(x=\frac{53}{60}\)
Nếu, \(2x+\left(-\frac{3}{2}\right)=-\frac{4}{15}\)
\(2x=\frac{37}{30}\)
\(x=\frac{37}{60}\)
Vậy \(x\in\text{{}\frac{53}{60};\frac{37}{60}\)}
b) \(\left|\frac{2}{7}x-\frac{1}{5}\right|-\left|-x+\frac{4}{9}\right|=0\)
\(\left|\frac{2}{7}x-\frac{1}{5}\right|=\left|-x+\frac{4}{9}\right|\)
\(\Rightarrow\left|\frac{2}{7}x-\frac{1}{5}\right|\in\text{{}-x+\frac{4}{9};-\left(x+\frac{4}{9}\right)\)}
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-x+\frac{4}{9}\)
\(x=\frac{203}{405}\)
Nếu, \(\frac{2}{7}x-\frac{1}{5}=-\left(-x+\frac{4}{9}\right)\)
\(\frac{2}{7}x-\frac{1}{5}=x-\frac{4}{9}\)
\(\frac{2}{7}x-x=\frac{1}{5}-\frac{4}{9}\)
\(-\frac{5}{7}x=-\frac{11}{45}\)
\(x=\frac{77}{225}\)
Vậy \(x\in\text{{}\frac{203}{405};\frac{77}{225}\)}
Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Bài làm :
\(a,\left(x-3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b,\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=\frac{1}{4}\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c,\left(\frac{1}{3}-x\right).\left(\frac{1}{2}+1:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1:x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d,\left(x+3\right)\left(x-4\right)+2\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-4+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Học tốt nhé
a, ( x - 3 ) . ( x - 4 ) = 0
=> x - 3 = 0 hoặc x - 4 = 0
Nếu x - 3 = 0 => x = 3
Nếu x - 4 = 0 => x = 4
b, (\(\frac{1}{2}\)x - 4 ) . ( x - \(\frac{1}{4}\)) = 0
=>( \(\frac{1}{2}\)x - 4 ) = 0 Hoặc ( x - \(\frac{1}{4}\)) = 0
Nếu ( \(\frac{1}{2}\)x - 4 ) = 0 => x = \(\frac{8}{1}\)
Nếu ( x - \(\frac{1}{4}\)) = 0 => x = \(\frac{1}{4}\)
c, (\(\frac{1}{3}\)- x ) . ( \(\frac{1}{2}\)+ 1 : x ) = 0
=> ( \(\frac{1}{3}\)- x ) = 0 Hoặc ( \(\frac{1}{2}\)+ 1 : x ) = 0
Nếu (\(\frac{1}{3}\)- x ) = 0 => x = \(\frac{1}{3}\)
Nếu ( \(\frac{1}{2}\)+ 1 : x ) = 0 => x = \(\frac{-2}{1}\)
d, ( x + 3 ) . ( x - 4 ) + 2.(x + 3 ) = 0
=> (X + 3 ) = 0 Hoặc ( x - 4 ) = 0 Hoặc 2. ( x + 3 ) = 0
Nếu x + 3 = 0 => x = 0
Nếu ( x - 4 ) = 0 => x = 4
Nếu 2.(x + 3) = 0 => x = 3
# Cụ MAIZ
a. ( x - 3 ) ( x - 4 ) = 0
<=> \(\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
b. \(\left(\frac{1}{2}x-4\right)\left(x-\frac{1}{4}\right)=0\)
<=> \(\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
Bài làm :
\(a\text{)}...\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}\)
\(b\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-4=0\\x-\frac{1}{4}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=4\\x=0+\frac{1}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=\frac{1}{4}\end{cases}}\)
\(c\text{)}...\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}-x=0\\\frac{1}{2}+1\div x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}-0\\1\div x=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}\)
\(d\text{)}...\Leftrightarrow\left(x+3\right)\left(x-4+2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
a) \(\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}:x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}:x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)
b)
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1+\frac{x+329}{5}+4=4\)
\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Mà \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
\(\Rightarrow x+329=0\)
\(\Rightarrow x=-329\)
Vậy \(x=-329\)
a: A=2/3*3/2*xy^2*x=x^2y^2
b: Bậc là 4
c: Khi x=-1 và y=2 thì A=(-1)^2*2^2=4
d: A=(xy)^2>0 khi x<>0 và y<>0
Lời giải:
a.
$x(x-\frac{1}{2})< 0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x< 0\\ x-\frac{1}{2}>0\end{matrix}\right.\\ \left\{\begin{matrix} x>0\\ x-\frac{1}{2}< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} 0> x>\frac{1}{2} (\text{vô lý})\\ 0< x< \frac{1}{2}(\text{chọn})\end{matrix}\right.\)
b.
\(\frac{x-1}{x+2}< 0\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x-1< 0\\ x+2>0\end{matrix}\right.\\ \left\{\begin{matrix} x-1>0\\ x+2< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} -2< x< 1(\text{chọn})\\ -2> x>1(\text{vô lý})\end{matrix}\right.\)