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\(\Leftrightarrow-5< x< -\dfrac{1037}{1260}\)
=>\(x\in\left\{-4;-3;-2;-1\right\}\)
\(1,\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2+y^2+z^2}{5}=\frac{x^2}{5}+\frac{y^2}{5}+\frac{z^2}{5}\)
\(=>\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}-\left(\frac{x^2}{5}+\frac{y^2}{5}+\frac{z^2}{5}\right)=0\)
\(=>\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)
\(=>\left(\frac{5x^2}{10}-\frac{2x^2}{10}\right)+\left(\frac{5y^2}{15}-\frac{3y^2}{15}\right)+\left(\frac{5z^2}{20}-\frac{4z^2}{20}\right)=0\)
\(=>\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)
Tổng 3 số không âm=0 <=> chúng đều=0
\(< =>\frac{3}{10}x^2=\frac{2}{15}y^2=\frac{1}{20}z^2=0< =>x=y=z=0\)
Vậy x=y=z=0
\(2,x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(=>x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}-4=0\)
\(=>\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)=0\)
\(=>\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
\(=>\left(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
\(=>\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
Tổng 2 số không âm=0 <=> chúng đều=0
\(< =>\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}< =>\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\end{cases}< =>\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}}}\)\(< =>\hept{\begin{cases}x\in\left\{-1;1\right\}\\y\in\left\{-1;1\right\}\end{cases}}\)
Vậy có 4 cặp (x;y) cần tìm là (1;1) ;(1;-1);(-1;1);(-1;-1)
Ta đặt cm là A
Vì 1/2 < 2/3 ; 3/4 < 4/5 ; 5/6 < 6/7 ; ...;99/100<100/101
=> A = 1/2 x 3/4 x 5/6 x...x 99/100 < B= 2/3 X 4/5 X 6/7 X....X100/101
=> A x A < A x B = 1 x 3 x 5 x 99 / 2 x 4 x 6 x ......x 100 x 2 x 4 x 6 x ...x 100/3 x 5 x 7 x ...x 101
Ta rút gọn 2 x 4 x 6 x ..x 100 và 3 x 5 x ...x 99 ta còn 1/101
=>A^2 < 1/101 => A^2 < 1/101 < 1/100 = > A ^ 2 <1/100 => A^2 ,(1/10 ^2
=> A < 1/10
Chứng minh A > 1/15
1/2 = 1/2
3/4 >2/3
5/6 > 4/5
......
99/100 > 98/99
A^2 > 1/2 x ( 1/2 x 2/3 x 3/4 x ...x 98/99 x 99/100
A^2 > 1/2 x 1/100
A^2 > 1/200 > 1/225
A^2 > (1/15) ^2
Vậy A > 1/15
\(\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}=\frac{x^2}{5}+\frac{y^2}{5}+\frac{z^2}{5}\)
\(\Rightarrow\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}-\frac{x^2}{5}-\frac{y^2}{5}-\frac{z^2}{5}=0\)
\(\Rightarrow\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)
\(\Rightarrow x^2\left(\frac{1}{2}-\frac{1}{5}\right)+y^2\left(\frac{1}{3}-\frac{1}{5}\right)+z^2\left(\frac{1}{4}-\frac{1}{5}\right)=0\)
Mà \(x^2\left(\frac{1}{2}-\frac{1}{5}\right)+y^2\left(\frac{1}{3}-\frac{1}{5}\right)+z^2\left(\frac{1}{4}-\frac{1}{5}\right)\ge0\)
Xảy ra khi \(\hept{\begin{cases}x^2\left(\frac{1}{2}-\frac{1}{5}\right)=0\\y^2\left(\frac{1}{3}-\frac{1}{5}\right)=0\\z^2\left(\frac{1}{4}-\frac{1}{5}\right)=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x^2=0\\y^2=0\\z^2=0\end{cases}}\)\(\Rightarrow x=y=z=0\)
Bài 1:
a) Ta có: \(2,3x-2\left(0,7+2x\right)=3,6-1,7x\)
\(\Leftrightarrow2,3x-1,4-4x-3,6+1,7x=0\)
\(\Leftrightarrow-5=0\)(vl)
Vậy: \(x\in\varnothing\)
b) Ta có: \(\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}\)
\(\Leftrightarrow\frac{4}{3}x=\frac{1}{2}+\frac{5}{6}=\frac{8}{6}=\frac{4}{3}\)
hay x=1
Vậy: x=1
c) Ta có: \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(\Leftrightarrow\frac{9x}{90}-\frac{3x}{90}-\frac{4x}{90}-\frac{72}{90}=0\)
\(\Leftrightarrow2x-72=0\)
\(\Leftrightarrow2\left(x-36\right)=0\)
mà 2>0
nên x-36=0
hay x=36
Vậy: x=36
d) Ta có: \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\Leftrightarrow12\left(10x+3\right)=8\left(7-8x\right)\)
\(\Leftrightarrow120x+36=56-64x\)
\(\Leftrightarrow120x+36-56+64x=0\)
\(\Leftrightarrow184x-20=0\)
\(\Leftrightarrow184x=20\)
hay \(x=\frac{5}{46}\)
Vậy: \(x=\frac{5}{46}\)
e) Ta có: \(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
\(\Leftrightarrow\frac{2\left(10x-5\right)}{36}+\frac{3\left(x+3\right)}{36}-\frac{6\left(7x+3\right)}{36}+\frac{4\left(12-x\right)}{36}=0\)
\(\Leftrightarrow2\left(10x-5\right)+3\left(x+3\right)-6\left(7x+3\right)+4\left(12-x\right)=0\)
\(\Leftrightarrow20x-10+3x+9-42x-18+48-4x=0\)
\(\Leftrightarrow-23x+29=0\)
\(\Leftrightarrow-23x=-29\)
hay \(x=\frac{29}{23}\)
Vậy: \(x=\frac{29}{23}\)
f) Ta có: \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
\(\Leftrightarrow\frac{2\left(x+4\right)}{10}-\frac{10x}{10}-\frac{50}{10}=\frac{25}{10}\)
\(\Leftrightarrow2x+8-10x-50-25=0\)
\(\Leftrightarrow-8x-67=0\)
\(\Leftrightarrow-8x=67\)
hay \(x=\frac{-67}{8}\)
Vậy: \(x=\frac{-67}{8}\)
g) Ta có: \(\frac{2-x}{4}=\frac{2\left(x+1\right)}{5}-\frac{3\left(2x-5\right)}{10}\)
\(\Leftrightarrow5\left(2-x\right)-8\left(x+1\right)+6\left(2x-5\right)=0\)
\(\Leftrightarrow10-5x-8x-8+12x-30=0\)
\(\Leftrightarrow-x-28=0\)
\(\Leftrightarrow-x=28\)
hay x=-28
Vậy: x=-28
h) Ta có: \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}+\frac{9\left(2x-1\right)}{12}-\frac{2\left(5x-3\right)}{12}-\frac{12x}{12}-\frac{5}{12}=0\)
\(\Leftrightarrow4x+8+18x-9-10x+6-12x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
Bài 2:
a) Ta có: \(5\left(x-1\right)\left(2x-1\right)=3\left(x+8\right)\left(x-1\right)\)
\(\Leftrightarrow5\left(x-1\right)\left(2x-1\right)-3\left(x-1\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(2x-1\right)-3\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(10x-5-3x-24\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x-29\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\7x-29=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\7x=29\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{29}{7}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{1;\frac{29}{7}\right\}\)
b) Ta có: \(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)(1)
Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+5\ge5\ne0\forall x\)(2)
Từ (1) và (2) suy ra:
\(\left[{}\begin{matrix}3x-2=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-6\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{\frac{2}{3};-6\right\}\)
c) Ta có: \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)-x+4=0\)
\(\Leftrightarrow27x^3-8-27x^3+1-x+4=0\)
\(\Leftrightarrow-x-3=0\)
\(\Leftrightarrow-x=3\)
hay x=-3
Vậy: Tập nghiệm S={-3}
d) Ta có: \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-\left(x^2+x-12\right)-5x=0\)
\(\Leftrightarrow x^2-x-x^2-x+12-5x=0\)
\(\Leftrightarrow12-7x=0\)
\(\Leftrightarrow7x=12\)
hay \(x=\frac{12}{7}\)
Vậy: Tập nghiệm \(S=\left\{\frac{12}{7}\right\}\)
e) Ta có: (2x+1)(2x-1)=4x(x-7)-3x
\(\Leftrightarrow4x^2-1-4x^2+28x+3x=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow31x=1\)
hay \(x=\frac{1}{31}\)
Vậy: Tập nghiệm \(S=\left\{\frac{1}{31}\right\}\)
\(a.\frac{x+5}{2021}+\frac{x+6}{2020}+\frac{x+7}{2019}=-3\\ \Leftrightarrow\frac{x+5}{2021}+1+\frac{x+6}{2020}+1+\frac{x+7}{2019}+1=0\\ \Leftrightarrow\frac{x+2026}{2021}+\frac{x+2026}{2020}+\frac{x+2026}{2019}=0\\ \Leftrightarrow\left(x+2026\right)\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\right)=0\\\Leftrightarrow x+2026=0\left(Vi\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\ne0\right)\\ \Leftrightarrow x=-2026\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2026\right\}\)
\(b.\frac{2-x}{100}-1=\frac{1-x}{101}-\frac{x}{102}\\ \Leftrightarrow\frac{2-x}{100}+1=\frac{1-x}{101}+1+1-\frac{x}{102}\\\Leftrightarrow \frac{102-x}{100}-\frac{102-x}{101}-\frac{102-x}{102}=0\\ \Leftrightarrow\left(102-x\right)\left(\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\right)=0\\ \Leftrightarrow102-x=0\left(Vi\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\ne0\right)\\ \Leftrightarrow x=102\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{102\right\}\)
c/ PT tương đương
\(\frac{x+1}{93}-1+\frac{x-2}{45}-2+\frac{x+4}{32}-3=0\)
\(\Leftrightarrow\frac{x-92}{93}+\frac{x-92}{45}+\frac{x-92}{32}=0\)
\(\Leftrightarrow\left(x-92\right)\left(\frac{1}{93}+\frac{1}{45}+\frac{1}{32}\right)=0\Rightarrow x=92\)
\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
<=> \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
<=> \(\frac{3\left(2x-1\right)}{5\cdot3}-\frac{5\left(x-2\right)}{3\cdot5}-\frac{x+7}{15}=0\)
<=> \(\frac{6x-3-5x+10-x-7}{15}=0\)
<=> \(\frac{-14}{15}=0\)
=> PT vô nghiệm
\(4_9^5:2_{18}^5-7
Xét VT, ta có:
\(4_9^5:2_{18}^5-7\)
\(\) \(=_9^{41}:_{18}^{41}-7\)
\(=2-7\)
\(=-5\)
Xét VP, ta có:
\(\left(3_5^1:3,2+4,5\cdot1_{45}^{31}\right):\left(-21_2^1\right)\)
\(=\left(\frac{16}{5}:\frac{16}{5}+\frac{45}{10}\cdot\frac{76}{45}\right):\left(-\frac{43}{2}\right)\)
\(=\left(1+\frac{76}{10}\right)\cdot\frac{-2}{43}\)
\(=\frac{86}{10}\cdot\frac{-2}{43}\)
\(\) \(=-\frac{4}{10}=-\frac25\)
Từ đó ta có: \(\) \(-5
\(mà\text{ }x\in Z\rarr x\in\left\lbrace-4;-3;-2;-1\right\rbrace\)
Có sai sót gì mong bạn thông cảm!
vế trái:
\(4\frac59:2\frac{5}{18}-7=\frac{41}{9}:\frac{41}{18}-7\)
\(=\frac{41}{9}\cdot\frac{18}{41}-7=2-7=-5\)
vế phải:
\(\left(3\frac15:3,2+4.5\cdot1\frac{31}{45}\right):\left(-21\frac12\right)\)
\(=\left(\frac{16}{5}:\frac{16}{5}+\frac92\cdot\frac{76}{45}\right):\left(-\frac{43}{2}\right)\)
\(=\left(1+\frac{38}{5}\right)\cdot\left(-\frac{2}{43}\right)=\frac{43}{5}\cdot\left(-\frac{2}{43}\right)=-\frac25\)
ta có: VT < x < VP
\(\Rightarrow-5
mà x là số nguyên nên \(x\in\left\lbrace-4;-3;-2;-1\right\rbrace\)