(1/3 + x) : 3 = 3/2
Mong giúp ạ
K
Khách
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NN
3
C
1 tháng 5 2022
\(\left(\dfrac{2}{5}-\dfrac{1}{4}\right):\dfrac{3}{4}x\dfrac{5}{2}=\dfrac{3}{20}:\dfrac{3}{4}x\dfrac{5}{2}=\dfrac{3}{20}x\dfrac{4}{3}x\dfrac{5}{2}=\dfrac{3x4x5}{20x3x2}=\dfrac{1}{2}\)
28 tháng 8 2021
a: Để A là số nguyên thì \(13⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;13;-13\right\}\)
hay \(x\in\left\{2;0;14;-12\right\}\)
NT
28 tháng 8 2021
b. Ta có \(B=\dfrac{x+3}{x-2}=\dfrac{x-2+3+2}{x-2}=1+\dfrac{5}{x-2}\)
Để \(B\) nhận giá trị nguyên thì\(5⋮\left(x-2\right)\Rightarrow\left(x-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\\x-2=5\\x-2=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=1\\\sqrt{x}=7\\\sqrt{x}=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\\x=49\end{matrix}\right.\)
Vậy tất cả các x thỏa mãn ycbt là x=9; x=1 hoặc x=49
14 tháng 5 2021
`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`
14 tháng 5 2021
1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)
\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)
\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)
\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)
\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)
\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)
DK
1
5 tháng 5 2022
ZnO - kẽm Oxit
KHSO4 - kali hidrosunfat
Mg(OH)2 - magie hidroxit
H3PO4 - axit photphoric
CaO - canxi oxit
HCl - axit clohidric
BaSO4 - bari sunfat
Na2CO3 - natri cacbonat
Zn(OH)2 - kẽm hidroxit
HP
Chứng minh rằng nếu x ≥ 2 thì: \(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}\) ≥ 2
Mong mn giúp đỡ.
1
14 tháng 6 2023
VT=|căn(x-2)+1|+|căn (x-2)-1|
=|căn (x-2)+1|+|1-căn x-2|>=|căn(x-2)+1+1-căn(x-2)|=2
MM
3
TC
10 tháng 5 2022
1)\(x=\left(\dfrac{1}{5}+\dfrac{2}{3}\right):\dfrac{4}{7}=\left(\dfrac{3}{15}+\dfrac{10}{15}\right):\dfrac{4}{7}=\dfrac{13}{15}:\dfrac{4}{7}=\dfrac{91}{60}\)
2)\(x=\left(-\dfrac{3}{10}-\dfrac{3}{4}\right):\dfrac{1}{2}=\left(-\dfrac{6}{20}-\dfrac{15}{20}\right):\dfrac{1}{2}=-\dfrac{21}{20}:\dfrac{1}{2}=-\dfrac{21}{10}\)
3)\(x=-\dfrac{24}{36}-\dfrac{7}{15}=-\dfrac{2}{3}-\dfrac{7}{15}=\dfrac{-10}{15}-\dfrac{7}{15}=-\dfrac{17}{15}\)
ID
10 tháng 5 2022
1) \(\dfrac{4}{7}x=\dfrac{1}{5}+\dfrac{2}{3}\)
\(\dfrac{4}{7}x=\dfrac{13}{15}\)
\(x=\dfrac{91}{60}\)
2) \(\dfrac{1}{2}x=-\dfrac{3}{10}-\dfrac{3}{4}\)
\(\dfrac{1}{2}x=-\dfrac{21}{20}\)
\(x=-\dfrac{21}{10}\)
3) \(x=-\dfrac{24}{36}-\dfrac{7}{15}\)
\(x=\dfrac{-2}{3}+\dfrac{7}{15}\)
\(x=-\dfrac{17}{15}\)
TM
25 tháng 4 2019
ĐKXĐ: x khác 1
\(M=\frac{x-\sqrt[3]{x}}{x-1}+\frac{1}{\sqrt[3]{x}-1}+\frac{1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}\)
\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+\sqrt[3]{x}+1+\sqrt[3]{x}-1}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{\left(\sqrt[3]{x}\right)^3-1}\)
\(=\frac{x-\sqrt[3]{x}}{x-1}+\frac{\sqrt[3]{x^2}+2\sqrt[3]{x}}{x-1}=\frac{x+\sqrt[3]{x^2}+\sqrt[3]{x}}{x-1}=\frac{\sqrt[3]{x}\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{\left(\sqrt[3]{x}-1\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}\)
\(=\frac{\sqrt[3]{x}}{\sqrt[3]{x}-1}\)
bạn nhớ kiểm tra lại nhé
1/3 + x = 3/2 x 3
1/3 + x = 9/2
x = 9/2 - 1/3 = 25/6
Kết quả là 25/6