giải phương trình
a) 8-x/x-7 - 8 = 1/x-7
b) x+5/x-5 - x-5/x+5 = 20/x^2 - 25
giải các phương trình sau;
a, 7x-3/x-1=2/3
b,2(3-7x)/1+x=1/2
c, 1/x-2 +3 = 3-x/x-2
d,8-x/x-7 -8 =1/x-7
e, x+5/x-5 - x-5/x+5 =20/x^2-25
f, 1/x-1 + 2/x+1 = x/x^2 -1
Giải các phương trình sau:
A) 1/x-2+3=3-x/x-2
B)8-x/x-7-8=-26/x-7
C)1/x-2+1/x-3=2/x-1
D)X^3-1/X=(x-2)(x+2)+11/x
E)x+5/x-5-x-5/x+5=20/x^2-25
F)x+7/x-3+4x-2/x-5=5
a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
=>1+3x-6=3-x
=>3x-5=3-x
=>4x=8
hay x=2(loại)
b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)
=>8-x-8x+56=-26
=>-9x+64=-26
=>-9x=-90
hay x=10(nhận)
c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\dfrac{x-3+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{2}{x-1}\)
\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)
\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)
=>-7x+10x=12-5
=>3x=7
hay x=7/3(nhận)
Bài 1. Giải các phương trình sau :
a) 7x - 35 = 0 b) 4x - x - 18 = 0
c) x - 6 = 8 - x d) 48 - 5x = 39 - 2x
Bài 2. Giải các phương trình sau :
a) 5x - 8 = 4x - 5 b) 4 - (x - 5) = 5(x - 3x)
c) 32 - 4(0,5y - 5) = 3y + 2 d) 2,5(y - 1) = 2,5y
Bài 3. Giải các phương trình sau :
a) \(\frac{3x-7}{5}=\frac{2x-1}{3}\)
b) \(\frac{4x-7}{12}- x=\frac{3x}{8}\)
Bài 4. Giải các phương trình sau :
a) \(\frac{5x-8}{3}=\frac{1-3x}{2}\)
b) \(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)
Bài 5. Giải các phương trình sau :
a) 6(x - 7) = 5(x + 2) + x b) 5x - 8 = 2(x - 4) + 3
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
có bị viết nhầm thì thông cảm nha!
la`thu'hai nga`y 19 nhe
1.giải các phương trình sau:
a, 3(2x+1)/4 - 5x+3/6 = 2x-1/3 - 3-x/4
b, 19/4 - 2(3x-5)/5 = 3-2x/10 - 3x-1/4
c, x-2*3/2+3 + x-3*5/3+5 + x-5*2/5+2 = 10
d, x-3/5*7 + x-5/3*7 + x-7/3*5 = 2(1/3 + 1/5 + 1/7)
2. giải các phương trình:
a, x-1/9 + x-2/8 = x-3/7 + x-4/6
b, (1/1*2 + 1/2*3 + 1/3*4 + ... + 1/9*10) (x-1) + 1/10x = x- 9/10
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)
\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)
\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)
\(< =>45x-90+50x-150+12x-60=0\)
\(< =>107x-300=0< =>x=\frac{300}{107}\)
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
Rút gọn biểu thức
a. 5√2x - 2√8 + 7√18 với x≥0
Giải phương trình
a.5√2x + 1 = 21
b. √4x + 20 - 3√5 + x + 7√9x + 45 = 20
Giải phương trình:
1) (3x-1)^2-5(2x+1)^2+96x-3)(2x+1)=(x-1)^2
2) (x+2)^3-(x-2)^3=12(x-1)-8
3) x-1/4-5-2x/9=3x-2/3
4) 25x-655/95-5(x-12)/209=[89-3x-2(x-13)/5]/11
5) 29-x/21+27-x/23+25-x/25+23-x/27=-4
6) x-69/30+x-67/32=x-63/36+x-61/38
7)x+117/19+x+4/28+x+3/57=0
8) 59-x/41+57-x/43+2=x-55?45+x-53/47-2
9) Cho phương trình: mx+x-m^2=2x-2 (x là ẩn). Tìm m để phương trình:
a) Có nghiệm duy nhất
b) Vô số nghiệm
c) Vô nghiệm
Giải Phương trình sau:
a:(x+1)/4-(x+2)/5+(x+4)/7-(x+5)/8+(x+7)/10-(x+9)/12=0
b:x/2004+(x+1)/2005+(x+2)/2006+(x+3)/2007=4
a) \(\frac{x+1}{4}-\frac{x+2}{5}+\frac{x+4}{7}-\frac{x+5}{8}+\frac{x+7}{10}-\frac{x+9}{12}=0\)
\(\Leftrightarrow\)\(\frac{x+1}{4}-1-\frac{x+2}{5}+1+\frac{x+4}{7}-1-\frac{x+5}{8}+1+\frac{x+7}{10}-1-\frac{x+9}{12}+1=0\)
\(\Leftrightarrow\)\(\frac{x-3}{4}-\frac{3-x}{5}+\frac{x-3}{7}-\frac{3-x}{8}+\frac{x+3}{10}-\frac{3-x}{12}=0\)
\(\Leftrightarrow\)\(\frac{x-3}{4}+\frac{x-3}{5}+\frac{x-3}{7}+\frac{x-3}{8}+\frac{x-3}{10}+\frac{x-3}{12}=0\)
\(\Leftrightarrow\)\(\left(x-3\right)\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\right)=0\)
Vì \(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\ne0\)
\(\Rightarrow\)\(x-3=0\)
\(\Leftrightarrow\)\(x=3\)
Vậy...
b) \(\frac{x}{2004}+\frac{x+1}{2005}+\frac{x+2}{2006}+\frac{x+3}{2007}=4\)
\(\Leftrightarrow\)\(\frac{x}{2004}-1+\frac{x+1}{2005}-1+\frac{x+2}{2006}-1+\frac{x+3}{2007}-1=0\)
\(\Leftrightarrow\)\(\frac{x-2004}{2004}+\frac{x-2004}{2005}+\frac{x-2004}{2006}+\frac{x-2004}{2007}=0\)
\(\Leftrightarrow\)\(\left(x-2004\right)\left(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\ne0\)
\(\Rightarrow\)\(x-2004=0\)
\(\Leftrightarrow\)\(x=2004\)
Vậy...
tìm x
a) X-2/7=1/5
b) 15/8-X=3/2
c) X x 7/5=7/20
d) 18/25 : X=9/5
a :17/35
b:54/16
c:140 /35
d:162/50
a, X-2/7=1/5
X = 1/5 + 2/7
X = 17/35
b,15/8-X=3/2
X = 15/8-3/2
X = -5/8
c, X x 7/5= 7/20
X = 7/20 : 7/5
X = 1/4
d, 18/25 : X= 9/5
X = 18/25 :9/5
X = 2/5
Nhớ k cho mình nhé! cảm ơn và chúc bạn học tốt :))