Bài 1: Tìm x, biết: a) \(1\frac{1}{2}x+\frac{-4}{5}=4\) b) \(\frac{2}{3}x+\frac{-1}{2}x=\frac{-5}{12}\) c) \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
d,
\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)
e,
\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)
\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)
\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)
Vậy không tồn tại $x$ thỏa mãn đề bài.
f,
\(\frac{2x-1}{2}=\frac{5+3x}{3}\Leftrightarrow 3(2x-1)=2(5+3x)\)
\(\Leftrightarrow 6x-3=10+6x\)
\(\Leftrightarrow 13=0\) (vô lý)
Vậy không tồn tại $x$ thỏa mãn đề bài.
a,
$0-|x+1|=5$
$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)
Do đó không tồn tại $x$ thỏa mãn điều kiện đề.
b,
\(2-|\frac{3}{4}-x|=\frac{7}{12}\)
\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)
\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)
c,
\(2|\frac{1}{2}x-\frac{1}{3}|-\frac{3}{2}=\frac{1}{4}\)
\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)
\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)
\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)
1 tìm x biết ;
a, 0-|x + 1| = 5
b, 2 - | \(\frac{3}{4}\)- x | = \(\frac{7}{12}\)
c, 2 | \(\frac{1}{2}\)x - \(\frac{1}{3}\)| - \(\frac{3}{2}\)= \(\frac{1}{4}\)
d, | x - \(\frac{1}{3}\)| = \(\frac{5}{6}\)
e, \(\frac{3}{4}\)- 2 | 2x - \(\frac{2}{3}\)| = 2
f, \(\frac{2x-1}{2}\)= \(\frac{5+3x}{3}\)
Tìm x, biết:
a) 60%x + 0,4x + x :3 =2
b)1-\(\left(5\frac{3}{8}+x-7\frac{5}{24}\right):\left(-16\frac{2}{3}\right)\)
c)\(3\frac{1}{4}x-\frac{7}{6}x=\frac{-5}{12}+1\frac{2}{3}\)
Bài 2: Tính:
a) A= \(\frac{-45.58-45.42}{2+4+6+...+16+18}\)
b)1-2-3+4+5-6-7+...+601-602-603+604
b) \(\frac{\left(140\frac{3}{7}-138\frac{5}{12}\right):18\frac{1}{6}}{0,002}\)
Bài 3: Cho A và B, biết:
A=\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\) và B= \(\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}+\frac{4}{195}\)
Hãy so sánh A & B
bài 2 tìm x
a,\(\frac{-2}{3}.x+\frac{1}{5}=\frac{3}{10}\)
b,\(\left|x\right|-\frac{3}{4}=\frac{5}{3}\)
c,\(\frac{2}{3}.x-\frac{1}{2}=\frac{1}{10}\)
d,\(\frac{3}{5}+\frac{4}{9}:x=\frac{2}{3}\)
e,\(\left|x+75\%\right|=2\frac{1}{5}\)
i,\(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2.x\right)=0\)
k,\(\frac{4}{7}.x-\frac{2}{3}=\frac{1}{5}\)
l,\(\frac{2}{3}.x-\frac{3}{2}.x=\frac{5}{12}\)
m,\(\left|2.x-\frac{1}{3}\right|+\frac{5}{6}=1\)
n,\(\frac{1}{3}-\frac{7}{8}.x=\frac{1}{3}\)
11,\(\frac{x+2}{5}=\frac{7}{12}-1\frac{1}{4}\)
12,\(\left(2\frac{4}{5}.x-50\right):\frac{2}{3}=51\)
13,\(\frac{2}{5}+\frac{3}{5}.\left(3.x-3,7\right)=-\frac{53}{10}\)
14,\(\frac{7}{9}:\left(2+\frac{3}{4}.x\right)+\frac{5}{9}=\frac{23}{27}\)
Bài 1 : Tìm x biết :
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
b, \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
c,\(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
Bài 2 : Tìm x biết :
a, | 2x - 5 | = x +1
b, | 3x - 2 | -1 = x
c, | 3x - 7 | = 2x + 1
d, | 2x-1 | +1 = x
1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)
=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)
b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c) TT
a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)
=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)
=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)
=> \(\left|50x-140\right|=\left|25x+24\right|\)
=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)
c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)
=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)
Bài 2 : a. |2x - 5| = x + 1
TH1 : 2x - 5 = x + 1
=> 2x - 5 - x = 1
=> 2x - x - 5 = 1
=> 2x - x = 6
=> x = 6
TH2 : -2x + 5 = x + 1
=> -2x + 5 - x = 1
=> -2x - x + 5 = 1
=> -3x = -4
=> x = 4/3
Ba bài còn lại tương tự
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
1
fddfssdfdsfdssssssssssssssffffffffffffffffffsssssssssssssssssssfsssssssssssssssssssssssfffffffffffffffEz lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
Bài 1:Thực hiện phép tính
\(a,75\%-1\frac{1}{2}+0,5:\frac{5}{12}-\left(\frac{-1}{2}\right)^2\) \(b,\left(\frac{5}{7}.0,6-5:3\frac{1}{2}\right).\left(40\%-1,4\right).\left(-2\right)^3\)
Bài 2: Tính nhanh
\(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
Bài 3:Tìm x biết
\(a,\frac{-2}{3}-\frac{1}{3}.\left(2x-5\right)=\frac{3}{2}\) \(b,\frac{2}{5}.x+\frac{1}{2}=\frac{-3}{4}\) \(\frac{1}{3}.x-8=\frac{1}{2}\) \(x-\frac{1}{4}=\frac{5}{8}.\frac{2}{3}\)
Bài 4: Tìm x biết
\(a,\left(\frac{3}{4}.x+2\frac{1}{2}\right).\frac{-2}{3}=\frac{1}{8}\) \(b,\frac{1}{3}.x-0,5=0,75\)
Bài 5: Tìm x biết
\(a,\frac{2}{5}+\frac{3}{5}.\left(3x-3,7\right)=-\frac{53}{10}\) \(b,5,2.x+7\frac{2}{5}=6\frac{3}{4}\) \(2,4:\left(\frac{-1}{2}-x\right)=1\frac{3}{5}\)
Bài 6:Tìm số tự nhiên x,biết: \(\left(x-5\right).\frac{30}{100}=\frac{20.x}{100}+5\)
Mik đang cần gấp giúp mik nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
bài 1. Tính giái trị các biểu thức là phân số :
a) \(M=\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}\)
b)\(N=\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}\)
\(\frac{ }{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)
c) \(H=x+\frac{0,2-0,375+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{22}vs,x=\frac{-1}{3}}\)
bài 2 : tìm x biết :
a)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
b)\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
c)\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
giúp mình với m.n
bài 15 tìm x biết
a\(\frac{x}{4}-\frac{3}{7}+\frac{2}{5}=\frac{31}{140}\)
b\(\frac{5}{12}+\frac{5}{x}-\frac{1}{8}=\frac{1}{2}\)
c\(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
d\(\frac{3}{4}-x+\frac{6}{-11}=\frac{5}{6}\)
e\(x-\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)