Tìm x
a,\(\left(\dfrac{3}{4}-5x\right)\)-\(\left(2x-4\dfrac{1}{4}\right)\)=\(6\dfrac{2}{7}\)
Help me, mai mình đi học
\(\dfrac{0,8:\left(\dfrac{4}{5}.1,25\right)}{0,64-\dfrac{1}{25}}\)+\(\dfrac{\left(100-\dfrac{2}{5}\right):\dfrac{4}{7}}{6.\left(\dfrac{5}{9}-3\dfrac{1}{4}\right).2\dfrac{2}{17}}\)
help me, please
Nhanh lên nha, mai mình đi học rùi
Thank you so much
Quy đồng, ta được:
\(\dfrac{0,8:\left(\dfrac{4}{5}.\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{500-2}{5}\right):\dfrac{4}{7}}{6\left(\dfrac{5}{9}-\dfrac{13}{4}\right).\dfrac{36}{17}}\)
\(=\dfrac{0,8:1}{\dfrac{15}{25}}+\dfrac{\dfrac{498}{5}.\dfrac{7}{4}}{6\left(\dfrac{20-117}{36}\right).\dfrac{36}{17}}\)
\(=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}+\dfrac{\dfrac{1743}{10}}{6.\dfrac{-97}{36}.\dfrac{36}{17}}\)
\(=\dfrac{4}{3}+\dfrac{\dfrac{1743}{10}}{\dfrac{-582}{17}}\)
\(=\dfrac{4}{3}-\dfrac{9877}{1940}=\dfrac{4.1940-9877.3}{3.1940}=\dfrac{-21871}{5820}\)
Số to quá !!!!
\(a.\dfrac{5x-20}{3}-\dfrac{x\left(1-3x\right)}{3}-\dfrac{5x}{4}\)
b.\(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
c.\(\dfrac{-1}{2x+3}< 0\)
CỨU MÌNH VỚI MỌI NGƯỜI MÌNH ĐANG CẦN GẤP MAI MINK KTR RỒI. HELP ME!!!!!!!!!!
a. có vấn đề
b.
\(\dfrac{5x^2-3x}{5}+\dfrac{3x+1}{4}< \dfrac{x\left(2x+1\right)}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow20x^2-12x+15x+5< 20x^2+10x-30\)
\(\Leftrightarrow-22x+5x< -30-5\)
\(\Leftrightarrow-17x< -35\)
\(\Leftrightarrow x>\dfrac{35}{17}\)
Tìm x biết:
\(a,\left(x-\dfrac{3}{4}\right)+50\%=\dfrac{1}{6}\)
\(b,\dfrac{1}{2}x-\dfrac{5}{6}x=\dfrac{7}{2}\)
\(c,\left(4-x\right)\left(3x+5\right)=0\)
\(d,\dfrac{x}{16}=\dfrac{50}{32}\)
\(e,\left(2x-3\right)+\dfrac{3}{2}=-\dfrac{1}{4}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
Giải các phương trình sau :
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
b,\(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
c,\(\dfrac{2x}{3}+\dfrac{3x-5}{4}=\dfrac{3\left(2x-1\right)}{2}-\dfrac{7}{6}\)
d,\(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)
e,\(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
a, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
⇔ \(6x^2-5x+3=2x-9x+6x^2\)
⇔ \(6x^2-5x+3-6x^2+9x-2x=0\)
⇔ \(2x+3=0\)
⇔ \(2x=-3\)
⇔ \(x=-\dfrac{3}{2}\)
b, \(\dfrac{2\left(x-4\right)}{4}-\dfrac{3+2x}{10}=x+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20\left(x-4\right)}{4.10}-\dfrac{4\left(3+2x\right)}{4.10}=\dfrac{5x}{5}+\dfrac{1-x}{5}\)
⇔ \(\dfrac{20x-80}{40}-\dfrac{12+8x}{40}=\dfrac{5x+1-x}{5}\)
⇔ \(\dfrac{20x-80-12-8x}{40}=\dfrac{4x+1}{5}\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{4x+1}{5}=0\)
⇔ \(\dfrac{12x-92}{40}-\dfrac{8\left(4x+1\right)}{40}=0\)
⇔ \(12x-92-8\left(4x+1\right)=0\)
⇔ 12x - 92 - 32x - 8 = 0
⇔ -100 - 20x = 0
⇔ 20x = -100
⇔ x = -100 : 20
⇔ x = -5
c, \(\dfrac{2x}{3}+\dfrac{3x-5}{4}=\dfrac{3\left(2x-1\right)}{2}-\dfrac{7}{6}\)
⇔ \(\dfrac{8x}{12}+\dfrac{9x-15}{12}=\dfrac{18x-9}{6}-\dfrac{7}{6}\)
⇔ \(\dfrac{17x-15}{12}=\dfrac{18x-16}{6}\)
⇔ \(\dfrac{17x-15}{12}-\dfrac{18x-16}{6}=0\)
⇔ \(\dfrac{17x-15}{12}-\dfrac{36x-32}{12}=0\)
⇔ 17x - 15 - 36 + 32 = 0
⇔ 17 - 19x = 0
⇔ 19x = 17
⇔ x = \(\dfrac{17}{19}\)
Giải các phương trình sau :
a) \(\dfrac{9x-0,7}{4}-\dfrac{5x-1,5}{7}=\dfrac{7x-1,1}{3}-\dfrac{5\left(0,4-2x\right)}{6}\)
b) \(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}=1-\dfrac{4}{\left(x-1\right)\left(x+3\right)}\)
c) \(\dfrac{3}{4\left(x-5\right)}+\dfrac{15}{50-2x^2}=-\dfrac{7}{6\left(x+5\right)}\)
d) \(\dfrac{8x^2}{3\left(1-4x^2\right)}=\dfrac{2x}{6x-3}-\dfrac{1+8x}{4+8x}\)
câu 1 \(\dfrac{3\left(2x+1\right)}{4}\)-\(\dfrac{2\left(3x-1\right)}{5}\)=5
câu 2 \(\dfrac{5x-3}{6}\)+5-\(\dfrac{7x-1}{4}\)=\(\dfrac{x+2}{12}\)
câu 3 \(\dfrac{\left(x-1\right)\left(x+1\right)}{2}\)+\(\dfrac{3\left(x+1\right)}{4}\)=\(\dfrac{\left(x-2\right)^2}{2}\)
câu 4 \(\dfrac{\left(x-4\right)^3}{6}\)+1=\(\dfrac{x\left(x+1\right)}{2}\)-\(\dfrac{\left(x-5\right)\left(x+5\right)}{3}\)
câu 5 \(\dfrac{3\left(x+2\right)^3}{5}\)-\(\dfrac{\left(x-1\right)^2}{10}\)=\(\dfrac{\left(x-3\right)\left(x+3\right)}{2}\)
-Mình xin lỗi.Nhưng thầy mình chưa giảng về bài này và làm bài tập với lại thầy bắt phải làm nên ko cho điểm xấu nên các bạn giúp mình được ko.Mình cầu xin sự giúp đỡ của các bạn và cảm ơn rất nhiều ( các bạn biết câu nào làm câu đó nha <:{ )
Câu 1:
=>15(2x+1)-8(3x-1)=100
=>30x+15-24x+8=100
=>6x+23=100
hay x=77/6
Câu 2:
=>2(5x-3)+12-3(7x-1)=x+2
=>10x-6+12-21x+3-x-2=0
=>-12x=-7
hay x=7/12
Câu 3:
\(\Leftrightarrow2\left(x^2-1\right)+3\left(x+1\right)=2\left(x^2-4x+4\right)\)
\(\Leftrightarrow2x^2-2+3x+3-2x^2+8x-8=0\)
=>11x-7=0
hay x=-7/11
Câu 4:
(x - 4)^3/6 + 1 = x(x + 1)/2 - (x - 5)(x + 5)/3
<=> (x - 4)^3 + 6/6 = x^2 + x/2 - x^2 - 25/3
<=> (x - 4)^3 + 6/6 = 3x^2 + 3x - 2x^2 + 50/6
<=> (x - 4)^3 + 6 = 3x^2 + 3x - 2x^2 + 50
<=> x^3 - 12x^2 + 48x - 58 = x^2 + 3x + 50
<=> x^3 -13x^2 + 45x - 108 = 0
Đến đây bạn bấm máy nhẩm nghiệm là ra nhé
Câu 5:
3(x + 2)^3/5 - (x - 1)^2/10 = (x - 3)(x + 3)/2
<=> 6(x + 2)^3 - (x - 1)^2/10 = 5(x^2 - 9)/10
<=> 6(x + 2)^3 - (x - 1)^2 = 5(x^2 - 9)
<=> 6x^3 + 36x^2 + 72x + 48 - x^2 + 2x - 1 - 5x^2 + 45 = 0
<=> 6x^3 + 30x^2 + 74x + 92 = 0
Đến đây bạn bấm máy nhẩm nghiệm như câu 4 nhé
Cho đa thức \(P\left(x\right)=\dfrac{5x^3}{4}+\dfrac{5x^2}{6}-\dfrac{21x}{4}+\dfrac{1}{6}\). Tìm số dư khi chia \(P\left(x\right)\) cho \(2x-5\).
Lời giải:
Theo định lý Bê-du về phép chia đa thức, số dư của $P(x)$ khi chia $2x-5$ là $P(\frac{5}{2})=\frac{5}{4}(\frac{5}{2})^3+\frac{5}{6}(\frac{5}{2})^2-\frac{21}{4}.\frac{5}{2}+\frac{1}{6}=\frac{377}{32}$
Giải phương trình
a) \(\dfrac{3}{5x-1}\)+ \(\dfrac{2}{3-5x}\)=\(\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
b) \(\dfrac{5-x}{4x^2-8x}\)+\(\dfrac{7}{8x}\)=\(\dfrac{x-1}{2x\left(x-2\right)}\)+\(\dfrac{1}{8x-16}\)
a:Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)
=>3x-9-10x+2=-4
=>-7x-7=-4
=>-7x=3
=>x=-3/7
b: =>\(\dfrac{5-x}{4x\left(x-2\right)}+\dfrac{7}{8x}=\dfrac{x-1}{2x\left(x-2\right)}+\dfrac{1}{8\left(x-2\right)}\)
=>\(2\left(5-x\right)+7\left(x-2\right)=4\left(x-1\right)+x\)
=>10-2x+7x-14=4x-4+x
=>5x-4=5x-4
=>0x=0(luôn đúng)
Vậy: S=R\{0;2}
a) \(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
b) \(\dfrac{\left(3x-1\right)\left(x+2\right)}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)
c) \(x-\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
d) \(\dfrac{x-1}{2}+\dfrac{x-1}{3}-\dfrac{x-1}{6}=2\)
a, \(\Rightarrow10x-4+6x=6+15-9x\Leftrightarrow7x=25\Leftrightarrow x=\dfrac{25}{7}\)
b, \(\Rightarrow2\left(3x^2+5x-2\right)-6x^2-3=33\Leftrightarrow10x-7=33\Leftrightarrow x=4\)
c, \(\Rightarrow12x-10x-4=21-9x\Leftrightarrow11x=25\Leftrightarrow x=\dfrac{25}{11}\)
d, \(\Rightarrow3x-3+2x-2-x+1=12\Leftrightarrow4x=16\Leftrightarrow x=4\)
\(\dfrac{5x-2}{3}+x=1+\dfrac{5-3x}{2}\)
\(\Leftrightarrow\dfrac{5x-2+3x}{3}=\dfrac{2+5-3x}{2}\)
\(\Leftrightarrow\dfrac{8x-2}{3}=\dfrac{7-3x}{2}\)
\(\Leftrightarrow16x-4=21-9x\)
\(\Leftrightarrow16x+9x=21+4\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
\(a,\dfrac{5x-2}{3}+x=1+\dfrac{-3x+5}{2}\)
\(2\left(5x-2\right)+6x=-9x+21\)
\(16x+9x=21+4\)
\(25x=25\)
\(x=1\)
\(b,\dfrac{3x^2+5x-2}{3}-\dfrac{2x^2+1}{2}=\dfrac{11}{2}\)
\(\dfrac{6x^2=10x-4-6x^2-3}{6}=\dfrac{11}{2}\)
\(\dfrac{10x-4-3}{6}=\dfrac{11}{2}\)
\(\dfrac{10x-7}{6}=\dfrac{11}{2}\)
\(10x=33+7\)
\(x=4\)