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\(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
Để í ngoặc \(\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]\)
\(\Leftrightarrow\left[\frac{6}{7}+-\frac{6}{7}\right]\)
\(\Leftrightarrow0\)
Vậy biểu thức \(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)có giá trị bằng 0
1.a) có: \(|x-\frac{3}{2}|,|x+1|,\left|x-2\right|\ge0\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(x\ge0\Rightarrow x-\frac{3}{2}\ge\frac{-3}{2}\Rightarrow\left|x-\frac{3}{2}\right|\ge\left|\frac{-3}{2}\right|=\frac{3}{2}\Rightarrow\left|x-\frac{3}{2}\right|=x-\frac{3}{2}\)
cmtt: \(|x-2|=x-2\)
\(\Rightarrow3x-\frac{3}{2}+1-2=4x\)
\(\Rightarrow3x-\frac{5}{2}=4x\)
\(\Rightarrow x=\frac{-5}{2}\left(ko,t/m\right)\)
Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)
a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1
<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5
<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3
<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5
<=> |x - 5/3|/6 = 9/5
<=> |x - 5/3| = 9/5.6
<=> |x - 5/3| = 54/5
<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5
x = 54/5 + 5/3 x = -54/5 - 5/3
x = 187/15 x = -137/15
b) 2.|3x + 1| = 1/3.|3x + 1| + 5
<=> 2.|3x + 1| - 1/3.|3x + 1| = 5
<=> 5/3.|3x + 1| = 5
<=> 5.|3x + 1| = 5.3
<=> 5.|3x + 1| = 15
<=> |3x + 1| = 15 : 5
<=> |3x + 1| = 3
3x + 1 = 3 hoặc 3x + 1 = -3
3x = 3 - 1 3x = -3 - 1
3x = 2 3x = -4
x = 2/3 x = -4/3
=> x = 2/3 hoặc x = -4/3
c) làm tương tự câu a) mình hơi lời
Làm câu c) cho
\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)
\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)
\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)
Giải tiếp nha
1) 22x + 1 = 32
=> 22x + 1 = 25
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 2
(2) 3.x3 - 100 = 275
=> 3x3 = 275 + 100
=> 3x3 = 375
=> x3 = 375 : 3
=> x3 = 125
=> x3 = 53
=> x = 5
(4) (x - 1)3 - 25 = 72
=> (x - 1)3 = 49 + 32
=> (x - 1)3 = 81
(xem lại đề)
5) Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{3}=\frac{y}{5}=\frac{x-y}{3-5}=\frac{-4}{-2}=2\)
=> \(\hept{\begin{cases}\frac{x}{3}=2\\\frac{y}{5}=2\end{cases}}\) => \(\hept{\begin{cases}x=2.3=6\\y=2.5=10\end{cases}}\)
Vậy ...
6) Ta có: \(\frac{x}{2}=\frac{y}{3}\) => \(\frac{x}{10}=\frac{y}{15}\)
\(\frac{y}{5}=\frac{z}{4}\) => \(\frac{y}{15}=\frac{z}{12}\)
=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=\frac{-49}{37}\)
=> \(\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}}\) => \(\hept{\begin{cases}x=-\frac{49}{37}\cdot10=\frac{-490}{37}\\y=-\frac{49}{37}\cdot15=-\frac{735}{37}\\z=-\frac{49}{37}\cdot12=-\frac{588}{37}\end{cases}}\)
Vậy ...
mk lm bài mà mk cho là ''khó'' nhất thôi nha
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và \(x+y+z=-49\)
\(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}\left(1\right)\)
\(\frac{y}{5}=\frac{z}{4}\Rightarrow\frac{y}{15}=\frac{z}{12}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)
ADTC dãy tỉ số bằng nhau ta có
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x+y+z}{10+15+12}=-\frac{49}{37}\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=-\frac{49}{37}\\\frac{y}{15}=-\frac{49}{37}\\\frac{z}{12}=-\frac{49}{37}\end{cases}\Rightarrow\hept{\begin{cases}x=-\frac{49}{37}.10=-\frac{490}{37}\\y=-\frac{49}{37}.15=-\frac{735}{37}\\z=-\frac{49}{37}.12=-\frac{588}{37}\end{cases}}}\)
Bài 1:
\(M\left(1\right)=a+b+6\)
Mà \(M\left(1\right)=0\)
\(\Rightarrow a+b+6=0\)
\(\Rightarrow a+b=-6\)( * )
\(\Rightarrow2a+2b=-12\) (1)
Ta có: \(M\left(-2\right)=4a-2b+6\)
Mà \(M\left(-2\right)=0\)
\(\Rightarrow4a-2b=-6\)(2)
Lấy (1) cộng (2) ta được:
\(6a=-18\)
\(a=-3\)
Thay a=-3 vào (* ) ta được:
\(b=-3\)
Vậy a=-3 ; b=-3
Bài 2:
a) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{8}-\frac{y}{4}=\frac{5}{x}\)
\(\Leftrightarrow\frac{1}{8}-\frac{2y}{8}=\frac{5}{x}\)
\(\Leftrightarrow\frac{1-2y}{8}=\frac{5}{x}\)
\(\Leftrightarrow\left(1-2y\right).x=5.8\)
\(\Leftrightarrow\left(1-2y\right).x=40\)
Vì \(x,y\in Z\Rightarrow1-2y\in Z\)
mà \(40=1.40=40.1=5.8=8.5=\left(-1\right).\left(-40\right)=\left(-40\right).\left(-1\right)=\left(-5\right).\left(-8\right)=\left(-8\right).\left(-5\right)\)
Thử từng TH
Câu 2:
Sửa đề; \(Q\left(x\right)=x^{99}-100x^{98}+100x^{97}-100x^{96}\)
x=99 nên x+1=100
\(Q\left(x\right)=x^{99}-x^{98}\left(x+1\right)+x^{97}\left(x+1\right)-x^{96}\left(x+1\right)\)
\(=x^{99}-x^{99}-x^{98}+x^{98}+x^{97}-x^{97}-x^{96}\)
\(=-x^{96}=-99^{96}\)
\(\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{100}\left(1+2+...+100\right)\)
\(=\frac{1+2}{2}+\frac{1+2+3}{3}+...+\frac{1+2+...+100}{100}\)
\(=\frac{\left(1+2\right).2:2}{2}+\frac{\left(1+2+3\right).3:2}{3}+...+\frac{\left(1+2+...+100\right).100:2}{100}\)
\(=\left(1+2\right):2+\left(1+2+3\right):2+....\left(1+2+...+100\right):2\)
\(=\left[\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+...+100\right)\right]:2\)
\(=\left(100.1+99.2+....+1.100\right):2=171700:2=85850\)
Nếu không hiểu cái trong ngoặc tính sao thì báo tớ ;)