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Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
-----------------------|-----------]|-/-/-/-/-/-/>
0 7
\(a,2x-6< 0\Leftrightarrow2x>6\Leftrightarrow x>3\)
\(b,5x+2x< 4+25\Leftrightarrow7x< 29\Leftrightarrow x< \frac{29}{7}\)
\(c,-5x+6>8-10+8x\Leftrightarrow-5x-8x>8-10-6\)
\(-13x>-8\Leftrightarrow x< \frac{8}{13}\)
\(d,3x-12\le2-4x\Leftrightarrow3x+4x\le2+12\)
\(\Leftrightarrow7x\le14\Leftrightarrow x\le2\)
\(e,\frac{3\left(x-3\right)}{6}>\frac{2\left(2x-5\right)}{6}+\frac{6}{6}\Rightarrow3x-9>4x-10+6\)
\(\Leftrightarrow3x-4x>-4+9\Leftrightarrow x>-5\)
\(f,3\left(2x-3\right)>1+2\left(2+2x\right)\Leftrightarrow6x-9>1+4+4x\)
\(6x-4x>14\Leftrightarrow2x>14\Leftrightarrow x>7\)
Tự biểu diễn nha!
17) \(ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1-3x^2-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow-4x^2+3x+1=0\)
\(\Leftrightarrow-\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-\frac{1}{4}\left(tm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{4}\right\}\)
18) \(ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
19) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\\x\ne\frac{1}{2}\end{cases}}\)
\(\frac{x+4}{2x^3-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\frac{x+4}{\left(2x-1\right)\left(x-2\right)}+\frac{x+1}{\left(2x-1\right)\left(x-3\right)}-\frac{2x+5}{\left(2x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-12+x^2-x-2-2x^2-x+10}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow x=-4\)(TM)
Vậy tập nghiệm của phương trình là \(S=\left\{-4\right\}\)
20) \(ĐKXĐ:x\ne0\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}-\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)-3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow x^4+x-x^4+x-3=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\)(TM)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{3}{2}\right\}\)
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
a) \(\left(x-5\right)^2+\left(x^2-25\right)=0\)
\(\Leftrightarrow\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+x+5\right)=0\)
\(\Leftrightarrow2x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
b) \(\frac{x-2}{4}+\frac{2x-3}{3}=\frac{x-18}{6}\)
\(\Rightarrow\frac{3x-6}{12}+\frac{8x-12}{12}=\frac{2x-36}{12}\)
\(\Rightarrow\frac{11x-18}{12}=\frac{2x-36}{12}\)
\(\Rightarrow11x-18=2x-36\)
\(\Rightarrow11x-2x=18-36\)
\(\Rightarrow9x=-18\Rightarrow x=-2\)
c) \(\frac{1}{x-3}+\frac{x-3}{x+3}=\frac{5x-6}{x^2-9}\)
\(\Rightarrow\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}=\frac{5x-6}{x^2-9}\)
\(\Rightarrow\frac{x+3}{\left(x+3\right)\left(x-3\right)}+\frac{x^2-6x+9}{\left(x+3\right)\left(x-3\right)}=\frac{5x-6}{x^2-9}\)
\(\Rightarrow\frac{x^2-5x+12}{x^2-9}=\frac{5x-6}{x^2-9}\)
\(\Rightarrow x^2-5x+12=5x-6\)
\(\Rightarrow x^2-10x+18=0\)
Giải biệt thức sẽ ra 2 nghiệm \(5+\sqrt{7}\)và \(5-\sqrt{7}\)
Gửi Cool: Lần sau đừng quên tìm điều kiện nhé. Câu c. ĐK: x khác 3 và x khác -3
A/ \(2\left(5x-3\right)=7x-18.\)
\(10x-6=7x-18\)
\(10-7x=6-18\)
\(3x=-12\)
\(x=-\frac{12}{3}=4\)
\(\Rightarrow S=\left\{4\right\}\)
B/ \(3x\left(x-2\right)+2x-4=0\)
\(3x\left(x-2\right)+2\left(x-2\right)=0\)
\(\left(x-2\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x-2=0\Rightarrow x=2\\3x+2=0\Rightarrow3x=-2\Rightarrow x=-\frac{2}{3}\end{cases}}\)
\(\Rightarrow S=\left\{2;-\frac{2}{3}\right\}\)
C/ \(\frac{x+2}{3}\frac{x-3}{2}=\frac{x+5}{4}\)
\(\frac{\left(x+2\right)\left(x-3\right)}{3.2}=\frac{x+5}{4}\)
\(\frac{x^2-3x+2x-6}{6}=\frac{x+5}{4}\)
\(\frac{x^2-x-6}{6}=\frac{x+5}{4}\)
\(\frac{2\left(x^2-x-6\right)}{12}=\frac{3\left(x+5\right)}{12}\)
\(\frac{2x^2-2x-12}{12}=\frac{3x+15}{12}\)
\(\Rightarrow2x^2-2x-12=3x+15\)
(chuyển vế r làm tiếp)
Bài 1 :
\(a,2\left(5x-3\right)=7x-18\)
\(\Leftrightarrow10x-6=7x-18\)
\(\Leftrightarrow10x-7x=6-18\)
\(\Leftrightarrow3x=-12\)
\(\Leftrightarrow x=-4\)
PT có nghiệm S = { -4 }
\(b,3x\left(x-2\right)+2x-4=0\)
\(\Leftrightarrow3x^2-6x+2x-4=0\)
\(\Leftrightarrow3x^2-4x-4=0\)
\(\Leftrightarrow3x^2-6x+2x-4=0\)
\(\Leftrightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x+2=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{3}\\x=2\end{cases}}\)
KL : ............
\(c,\frac{x+2}{3}-\frac{x-3}{2}=\frac{x+5}{4}\)
\(\Leftrightarrow\frac{4\left(x+2\right)}{12}-\frac{6\left(x-3\right)}{12}=\frac{3\left(x+5\right)}{12}\)
\(\Leftrightarrow4x+8-6x+18=3x+15\)
\(\Leftrightarrow4x-6x-3x=-8-18+15\)
\(\Leftrightarrow x=-9\)
KL : .......
c/ \(\frac{5}{x-3}+\frac{2}{x-2}=\frac{5}{\left(x-3\right)\left(x-2\right)}.\)
\(\frac{5\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}+\frac{2\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)
\(\frac{5x-10}{\left(x-3\right)\left(x-2\right)}+\frac{2x-6}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)
\(\frac{5x-10+2x-6}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)
\(\frac{7x-16}{\left(x-3\right)\left(x-2\right)}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)
\(\Rightarrow7x-16=5\)
\(\Rightarrow7x=5+16\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=\frac{21}{7}=3\)
\(\Rightarrow S=\left\{3\right\}\)
D/ \(\frac{7x-2}{3}-2x< 5-\frac{x-2}{4}\)
\(\frac{7x-2}{3}-\frac{2x.3}{3}< \frac{5.4}{4}-\frac{x-2}{4}\)
\(\frac{7x-2}{3}-\frac{6x}{3}< \frac{20}{4}-\frac{x-2}{4}\)
\(\frac{7x-2-6x}{3}< \frac{20-x+2}{4}\)
\(\frac{x-2}{3}< \frac{22-x}{4}\)
\(\frac{4\left(x-2\right)}{12}< \frac{3\left(22-x\right)}{12}\)
\(\frac{4x-8}{12}< \frac{66-3x}{12}\)
\(\Rightarrow4x-8=66-3x\)
\(\Rightarrow4x+3x=66+8\)
\(\Rightarrow7x=74\)
\(\Rightarrow x=\frac{74}{7}\)
\(\Rightarrow S=\left\{\frac{74}{7}\right\}\)
\(d,\frac{5}{x-3}+\frac{2}{x-2}=\frac{5}{\left(x-3\right)\left(x-2\right)}\)ĐKXĐ : \(x\ne3;2\)
\(\Leftrightarrow\frac{5\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}+\frac{2\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)
\(\Leftrightarrow5x-10+2x-6=5\)
\(\Leftrightarrow5x+2x=10+6+5\)
\(\Leftrightarrow7x=21\Leftrightarrow x=3\)
KL : PT có nghiệm S ={ 3}
\(e,\frac{4}{x}-\frac{5}{x+3}=1\)
\(\Leftrightarrow\frac{4\left(x+3\right)}{x\left(x+3\right)}-\frac{5x}{x\left(x+3\right)}=\frac{x\left(x+3\right)}{x\left(x+3\right)}\)
\(\Leftrightarrow4x+12-5x=x^2+3x\)
\(\Leftrightarrow\left(4x-5x-3x\right)+12-x^2=0\)
\(\Leftrightarrow12-4x-x^2=0\)
\(\Leftrightarrow12-6x+2x-x^2=0\)
\(\Leftrightarrow6\left(2-x\right)+x\left(2-x\right)=0\)
\(\Leftrightarrow\left(6+x\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}6+x=0\\2-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-6\\x=2\end{cases}}\)
KL : .....
Bài 3
A/ \(a+3va`b+3\)
Có \(a>b\)
\(\Rightarrow a+3>b+3\)
\(2a+5va`2b-1\)
Có \(a>b\Rightarrow2a>2b\)
\(5>\left(-1\right)\)
\(\Rightarrow2a+5>2b-1\)
C/\(4-7a va`4-7b\)
Có \(a>b\Rightarrow7a>7b\)
\(\Rightarrow4-7a>4-7b\)
Bài 2 ( bài này trục số bạn tự vẽ nha)
A/ \(5x-7>3x-15\)
\(5x-3x>7-15\)
\(2x>-8\)
\(x>-4\)
\(\Rightarrow S=\left\{x\in R;x>-4\right\}\)
B/ \(2x-5< 4x+1\)
\(2x-4x< 5+1\)
\(-2x< 6\)
\(x>-3\)
\(\Rightarrow S=\left\{x\in R;x>-3\right\}\)
c/ \(\frac{x+3}{5}-\frac{x+1}{3}>2\)
\(\frac{3\left(x+3\right)}{15}-\frac{5\left(x+1\right)}{15}>2\)
\(\frac{3x+9-5x-5}{15}>2\)
\(\frac{4-2x}{15}>\frac{30}{15}\)
\(\Rightarrow4-2x>30\)
\(\Rightarrow-2x>30-4\)
\(\Rightarrow-2x>26\)
\(\Rightarrow-x< -13\)
\(\Rightarrow S=\left\{x\in R;x< -13\right\}\)
Bài 2 :
\(a,5x-7>3x-15\)
\(\Leftrightarrow5x-3x>7-15\)
\(\Leftrightarrow2x>-8\)
\(\Leftrightarrow x>-4\)
b, \(b,2x-5< 4x+1\)
\(\Leftrightarrow2x-4x< 5+1\)
\(\Leftrightarrow-2x< 6\)
\(\Leftrightarrow x>-3\)
\(c,\frac{x+3}{5}-\frac{x+1}{3}>2\)
\(\Leftrightarrow\frac{3\left(x+3\right)}{15}-\frac{5\left(x+1\right)}{15}>\frac{30}{15}\)
\(\Leftrightarrow\frac{3x+9-5x-5}{15}>\frac{30}{15}\)
\(\Leftrightarrow4-2x>30\)
\(\Leftrightarrow-2x>30-4\)
\(\Leftrightarrow-2x>28\)
\(\Leftrightarrow x< -14\)
Cho mik sửa câu d bài 1 nhé cậu :> vì ta có
\(ĐKXĐ:\orbr{\begin{cases}x-2\ne0\Rightarrow x\ne2\\x-3\ne0\Rightarrow x\ne3\end{cases}}\)
mak mik tính được x = 3
\(\Rightarrow S=\varnothing\)
Sửa :
\(4-2x>30\)
\(\Leftrightarrow-2x>26\)
\(\Leftrightarrow x< -13\)
KL : ......
Bài 4 bài này vẽ sơ đồ thì rất dễ nhìn :v nhưng cậu tự vẽ nhé :3
Gọi v là vận tốc mak người B đi được (km/h)
Gọi v + 3 là vận tốc mak người A đi được (km/h)
Quãng đường mak người B đi được sau 2 giờ là 2v
Quãng đường mak người B đi được sau 2 giờ là 2 (v+3)
Ta có \(2v+2\left(v+3\right)=42\)
\(\Rightarrow2v+2v+6=42\)
\(\Rightarrow4v+6=42\)
\(\Rightarrow4v=42-6\)
\(\Rightarrow4v=36\)
\(\Rightarrow v=\frac{36}{4}=9\)(km/h)
=> Vận tốc người B đi được là 9km/h
=> Vận tốc người A đi được là \(v+3\Rightarrow9+3=12\)(km/h)
\(d,\frac{7x-2}{3}-2x< 5-\frac{x-2}{4}\)
\(\Leftrightarrow\frac{4\left(7x-2\right)}{12}-\frac{24x}{12}< \frac{60}{12}-\frac{3\left(x-2\right)}{12}\)
\(\Leftrightarrow28x-8-24x< 60-3x+6\)
\(\Leftrightarrow28x-24x-3x< 8+60+6\)
\(\Leftrightarrow x< 74\)
KL : ......
còn câu e bài 1 :v ngứa tay làm nốt
\(\frac{4}{x}-\frac{5}{x+3}=1\)
\(ĐKXĐ:\orbr{\begin{cases}x\ne0\\x+3\ne0\Rightarrow x\ne-3\end{cases}}\)
\(\frac{4}{x}-\frac{5}{x+3}=1\)
\(\frac{4\left(x+3\right)}{x\left(x+3\right)}-\frac{5x}{x+3}=1\)
\(\frac{4x+12}{x\left(x+3\right)}-\frac{5x}{x+3}=1\)
\(\frac{4x+12-5x}{x^2+3x}=1\)
\(\frac{12-x}{x^2+3x}=\frac{x^2+3x}{x^2+3x}\)
\(\Rightarrow12-x=x^2+3x\)
\(\Rightarrow12-x-3x-x^2=0\)
\(\Rightarrow12-4x-x^2=0\)
\(\Rightarrow-x^2-4x+12=0\)
\(\Rightarrow-x^2+2x-6x+12=0\)
\(\Rightarrow-\left(x^2-2x\right)-\left(6x-12\right)=0\)
\(\Rightarrow-x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(-x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}-x-6=0\Rightarrow-x=6\Rightarrow x=-6\\x-2=0\Rightarrow x=2\end{cases}}\)
\(\Rightarrow S=\left\{-6;2\right\}\)