Tìm x, biết
a) \(10x+3920=29000\)
b)\(\frac{x}{8}+2930=19202\)
c)\(x^2+x^4=18\)
d)\(x+x+x+304=604\)
e)\(x+5\frac{4}{5}=\frac{109}{5}\)
bài 1 giải phương trình
a) (2x+3)\(^2\)-3(x-4)(x+4)=\(\left(x-2\right)^2\)+1
b)(3x-2) (9x\(^2\)+6x+4)-(3x-1) (9x\(^2\)+3x+1)=x-4
c)x (x-1) -(x-3) (x+4)=5x
d) (2x+1)(2x-1)=4x(x-7)-3x
bài 2 giải phương trình
a)\(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
b)\(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}+\frac{12-x}{9}\)
c)\(\frac{10x+3}{8}=\frac{7-8x}{12}\)
d)\(\frac{x+4}{5}-x-5=\frac{x+3}{3}-\frac{x-2}{2}\)
Giải các phương trình sau
a, 5-(x-6)=4(3-2x)
b, 3-4x(25-2x)=8x2+x-300
c, x-\(\frac{2x-5}{5}+\frac{x+8}{6}=7+\frac{x-1}{3}\)
d, \(\frac{x+1}{15}+\frac{x+2}{7}\frac{x+4}{4}+6=0\)
e,\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)
g, \(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)
Các bạn giúp mk nha
\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
\(b)3-4x\left(25-2x\right)=8x^2+x-300\)
\(\Leftrightarrow3-100x+8x^2=8x^2+x-300\)
\(\Leftrightarrow101x-303=0\)
\(\Leftrightarrow101\left(x-3\right)=0\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Tìm x,biết
a)\(|x+\frac{1}{3}|=0\)
b)\(|x+\frac{3}{4}=\frac{1}{2}|\)
c)\(|\frac{5}{18}-x|-\frac{7}{24}=0\)
d)\(\frac{2}{5}-|\frac{1}{2}-x|=6\)
e)\(|\frac{3}{8}-x|+\frac{5}{6}=\frac{7}{4}\)
a, \(\left|x+\frac{1}{3}\right|=0\Leftrightarrow x=-\frac{1}{3}\)
b, \(\left|\frac{5}{18}-x\right|-\frac{7}{24}=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)
c, \(\frac{2}{5}-\left|\frac{1}{2}-x\right|=6\Leftrightarrow\left|\frac{1}{2}-x\right|=-\frac{28}{5}\)vô lí
Vì \(\left|\frac{1}{2}-x\right|\ge0\forall x\)*luôn dương* Mà \(-\frac{28}{5}< 0\)
=> Ko có x thỏa mãn
\(|x+\frac{1}{3}|=0\)
\(< =>x+\frac{1}{3}=0< =>x=-\frac{1}{3}\)
\(|x+\frac{3}{4}|=\frac{1}{2}\)
\(< =>\orbr{\begin{cases}x+\frac{3}{4}=\frac{1}{2}\\x+\frac{3}{4}=-\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{5}{4}\end{cases}}\)
\(|\frac{5}{18}-x|-\frac{7}{24}=0\)
\(< =>|\frac{5}{18}-x|=\frac{7}{24}\)
\(< =>\orbr{\begin{cases}\frac{5}{18}-x=\frac{7}{24}\\\frac{5}{18}-x=-\frac{7}{24}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=-\frac{1}{72}\\x=\frac{41}{72}\end{cases}}\)
\(\frac{2}{5}-|\frac{1}{2}-x|=6\)
\(< =>\frac{2}{5}-6=|\frac{1}{2}-x|\)
\(< =>\orbr{\begin{cases}\frac{1}{2}-x=-\frac{28}{5}\\\frac{1}{2}-x=\frac{28}{5}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{61}{10}\\x=-\frac{51}{10}\end{cases}}\)
Giải các phương trình sau:
a/ \(\frac{4x-3}{x-5}=\frac{29}{3}\)
b/ \(\frac{4x-5}{x-1}=2+\frac{x}{x-1}\)
c/ \(\frac{2x+5}{2x}-\frac{x}{x+5}=0\)
d/ \(\frac{2x-1}{5-3x}=2\)
e/ \(\frac{7}{x+2}=\frac{3}{x-5}\)
f/ \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
f, \(\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\)
\(\Leftrightarrow\) \(\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{2\left(10x-4\right)\left(11x-4\right)}{18\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\)
\(\Leftrightarrow\) 18(12x + 1) + 2(10x - 4)(11x - 4) = (20x + 17)(11x - 4)
\(\Leftrightarrow\) 216x + 18 + 220x2 − 168x + 32 = 220x2 + 107x − 68
\(\Leftrightarrow\) 216x + 18 + 220x2 − 168x + 32 - 220x2 - 107x + 68 = 0
\(\Leftrightarrow\) −59x + 118 = 0
\(\Leftrightarrow\) -59x = -118
\(\Leftrightarrow\) x = 2
Vậy S = {2}
Chúc bạn học tốt!
a) \(\frac{4x-3}{x-5}\)=\(\frac{29}{3}\) (ĐKXĐ:x≠5)
⇔\(\frac{3\left(4x-3\right)}{3\left(x-5\right)}\)=\(\frac{29\left(x-5\right)}{3\left(x-5\right)}\)
⇒12x-9=29x-145
⇔12x-9-29x+145=0
⇔-17x+136=0
⇔-17x=-136
⇔x=8
Vậy tập nghiệm của phương trình đã cho là:S={8}
GIẢI PHƯƠNG TRÌNH
a) (3x - 2) (9x2 + 6x + 4) - (3x - 1) (9x2 - 3x + 1) = x - 4
b) \(\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
c)\(\frac{10x-5}{18}+\frac{x+3}{12}=\frac{7x+3}{6}-\frac{12-x}{9}\)
d) \(\frac{10x+3}{8}=\frac{7-8x}{12}\)
e) \(\frac{x+4}{5}-x-5=\frac{x+3}{2}-\frac{x-2}{2}\)
Mng giúp Bơ nhaeee
Hứa đáp pưn đầy đủ ạaa
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
a.\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
b.\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)
c.\(\frac{12}{8-x^3}=1+\frac{1}{x+2}\)
d.\(\frac{x+25}{2x^2-50}-\frac{x+5}{x^2-5x}=\frac{5-x}{2x^2+10x}\)
e.\(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)
\(a.\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\left(dkxd:x\ne\pm1\right)\\\Leftrightarrow \frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}=\frac{16}{x^2-1}\\\Leftrightarrow \left(x+1\right)^2-\left(x-1\right)^2=16\\\Leftrightarrow \left(x+1-x+1\right)\left(x+1+x-1\right)-16=0\\\Leftrightarrow 4x-16=0\\\Leftrightarrow 4\left(x-4\right)=0\\\Leftrightarrow x-4=0\\ \Leftrightarrow x=4\left(tmdk\right)\)
\(b.\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(dkxd:x\ne\pm2\right)\\ \Leftrightarrow\frac{12}{x^2-4}-\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}+\frac{\left(x+7\right)\left(x-2\right)}{x^2-4}=0\\\Leftrightarrow 12-x^2-3x-2+x^2+5x-14=0\\ \Leftrightarrow2x-4=0\\\Leftrightarrow 2\left(x-2\right)=0\\\Leftrightarrow x-2=0\\\Leftrightarrow x=2\left(ktmdk\right)\)
Vô nghiệm
tìm x biết
a,\(\left(152\frac{2}{4}-148\frac{3}{8}\right):0,2=x:0,3\)
b,\(\left(85\frac{7}{30}-83\frac{5}{18}\right):2\frac{2}{3}=0,01x:4\)
c,\(\frac{x-1}{x+5}=\frac{6}{7}\)
d,\(\frac{x^2}{6}=\frac{24}{25}\)
e,\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)
g\(\frac{x-3}{x+5}=\frac{5}{7}\),
a, ( 152 +và 2/4 - 148 và 3/8 ) : 0,2 = x : 0,3
=> 33/8 : 1/5 = x : 3/10
=> x : 3/10 = 165/8
=> x = 99/10
b, ( 85 và 7/30 - 83 và 5/18 ) : 2 và 2/3 = 0,01x : 4
=> 88/45 : 8/3 = 0,01x : 4
=> 0,01x : 4 = 11/15
=> 0,01x = 44/15
=> x = 880/3
c, x - 1/ x + 5 = 6/7
=> 7( x - 1 ) = 6( x + 5 )
=> 7x - 7 = 6x + 30
=> 7x - 6x = 7 + 30
=> x = 37
d, x2/6 = 24/25
=> x2. 25 = 6 . 24
=> x2.25 = 144
=> x2 = 144/25
=> x = ( 12/5)2 hoặc x = ( -12/5)
g, x - 3/ x + 5 = 5/7
=> 7( x - 3 ) = 5 ( x + 5 )
=> 7x - 21 = 5x + 25
=> 7x - 5x = 21 + 25
=> 2x = 46
=> x = 23
Bài 1:
a)Tính tổng và tính tích các số nguyên x biết: \(x^2\)-15\(\le\)16
b)Tìm tất cả các số nguyên x biết: (|x|-3).(\(x^2\)+4)<0
Bài 2: Tìm các số nguyên x biết:
a)(x-3).(2x-5)=6
b)(x-1).(x+4)<0
c)\(5^{x+2}\)-\(5^{x-1}\)=3100
d)\(3^{x+1}\)-\(3^{x-2}\)=702
Bài 3:Tìm số nguyên x biết:
a)\(\frac{-8}{x}\)=\(\frac{-x}{18}\)
b)\(\frac{x+1}{22}=\frac{6}{x}\)
c)\(\frac{2x-1}{2}=\frac{5}{x}\)
d)\(\frac{2x-1}{21}=\frac{3}{2x+1}\)
e)\(\frac{10x+5}{6}=\frac{5}{x+1}\)
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Cách gọn:
1b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\)trái dấu
Mà \(x^2+4\ge0\) nên \(\left|x\right|-3< 0\Leftrightarrow\left|x\right|< 3\)
\(\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
bài 1:giải các phương trình sau:
a/2,3x-2.(0,7+2x)=3,6-1,7x
b/\(\frac{4}{3}\)x-\(\frac{5}{6}\)=\(\frac{1}{2}\)
c/\(\frac{x}{10}\)-(\(\frac{x}{30}\)+\(\frac{2x}{45}\))=\(\frac{4}{5}\)
d/\(\frac{10x+3}{8}\)=\(\frac{7-8x}{12}\)
e/\(\frac{10x-5}{18}\)+\(\frac{x+3}{12}\)=\(\frac{7x+3}{6}\)-\(\frac{12-x}{9}\)
f/\(\frac{x+4}{5}\)-x-5=\(\frac{x+3}{2}\)-\(\frac{x-2}{2}\)
g/\(\frac{2-x}{4}\)=\(\frac{2.\left(x+1\right)}{5}\)-\(\frac{3.\left(2x-5\right)}{10}\)
h/\(\frac{x+2}{3}\)+\(\frac{3.\left(2x-1\right)}{4}\)-\(\frac{5x-3}{6}\)=x+\(\frac{5}{12}\)
bài 2:giải các phương trình sau:
a/5.(x-1).(2x-1)=3.(x+8).(x-1)
b/(3x-2).(x+6).(\(^{x^2}\)+5)=0
c/(3x-2).(9\(^{x^2}\)+6x+4)-(3x-1).(9\(^{x^2}\)-3x+1)=x-4
d/x.(x-1)-(x-3).(x+4)=5x
e/(2x+1).(2x-1)=4x.(x-7)-3x
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 \(S=\left\{\frac{1}{31}\right\}\)