Giải các bất phương trình :
a) \(\dfrac{x^3-4\cdot x^2+5\cdot x-20}{x^3-x^2-10\cdot x-8}>0\)
b) \(\dfrac{x^2+2\cdot x+2}{x+1}>\dfrac{x^2+4\cdot x+5}{x+2}-1\)
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Giải các phương trình sau
a) -x^2+4cdot x+12cdotsqrt{2cdot x+1}
b) x+sqrt{x+dfrac{1}{2}+sqrt{x+dfrac{1}{4}}}2
c) 5cdot x^2-2cdot x+1left(4cdot x-1right)cdotsqrt{x^2+1}
d) left(2cdot x-1right)cdotsqrt{10-4cdot x^2}5-2cdot x
e) sqrt{2cdot x-1}-sqrt{x+1}2cdot x-4
f) sqrt{x^2-2cdot x}+sqrt{2cdot x^2+4cdot x}2cdot x
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Giải các phương trình sau
a) \(-x^2+4\cdot x+1=2\cdot\sqrt{2\cdot x+1}\)
b) \(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
c) \(5\cdot x^2-2\cdot x+1=\left(4\cdot x-1\right)\cdot\sqrt{x^2+1}\)
d) \(\left(2\cdot x-1\right)\cdot\sqrt{10-4\cdot x^2}=5-2\cdot x\)
e) \(\sqrt{2\cdot x-1}-\sqrt{x+1}=2\cdot x-4\)
f) \(\sqrt{x^2-2\cdot x}+\sqrt{2\cdot x^2+4\cdot x}=2\cdot x\)
câu b đk x>= -1/4
\(x+\sqrt{x+\dfrac{1}{2}+\sqrt{x+\dfrac{1}{4}}}=2\)
\(x+\sqrt{\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2}=2\)
\(\left(\sqrt{x+\dfrac{1}{4}}+\dfrac{1}{2}\right)^2=2\)
\(x+\dfrac{1}{4}=\left(\sqrt{2}-\dfrac{1}{2}\right)^2\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(x=\left(\sqrt{2}-\dfrac{1}{2}-\dfrac{1}{2}\right)\left(\sqrt{2}-\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(x=\sqrt{2}\left(\sqrt{2}-1\right)=2-\sqrt{2}\)
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Giải phương trình
\(1,\dfrac{x^2-2x-3}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-4x+6}{x-2}+\dfrac{x^2-6x+12}{x-3}\)
\(2,\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot[1+\dfrac{1}{x\cdot\left(x+2\right)}]=\dfrac{31}{16}\left(x\in N\right)\)
1,Tìm x, biết:
a/\(^{\left(x-1\right)^{x+2}}=^{\left(x-1\right)^{x+4}}\)
b/\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
c/\(4^x+3^x=2^x+6^x\left(STN\right)\)
GIÚP MÌNH NHANH NHA MÌNH TICK CHO
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
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SGK Toán 8 tập 1 – Chân trời sáng tạo
23 tháng 7 2023 lúc 15:40
Thực hiện các phép tính sau:a) dfrac{{8y}}{{3{x^2}}} cdot dfrac{{9{x^2}}}{{4{y^2}}}b) dfrac{{3x + {x^2}}}{{{x^2} + x + 1}} cdot dfrac{{3{x^3} - 3}}{{x + 3}}c) dfrac{{2{x^2} + 4}}{{x - 3}} cdot dfrac{{3x + 1}}{{x - 1}}:dfrac{{{x^2} + 2}}{{6 - 2x}}d) dfrac{{2{x^2}}}{{3{y^3}}}:left( { - dfrac{{4{x^3}}}{{21{y^2}}}} right)e) dfrac{{2x + 10}}{{{x^3} - 64}}:dfrac{{{{left( {x + 5} right)}^2}}}{{2x - 8}}f) dfrac{1}{{x + y}}left( {dfrac{{x + y}}{{xy}} - x - y} right) - dfrac{1}{{{x^2}}}:dfrac{y}{x}
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Thực hiện các phép tính sau:
a) \(\dfrac{{8y}}{{3{x^2}}} \cdot \dfrac{{9{x^2}}}{{4{y^2}}}\)
b) \(\dfrac{{3x + {x^2}}}{{{x^2} + x + 1}} \cdot \dfrac{{3{x^3} - 3}}{{x + 3}}\)
c) \(\dfrac{{2{x^2} + 4}}{{x - 3}} \cdot \dfrac{{3x + 1}}{{x - 1}}:\dfrac{{{x^2} + 2}}{{6 - 2x}}\)
d) \(\dfrac{{2{x^2}}}{{3{y^3}}}:\left( { - \dfrac{{4{x^3}}}{{21{y^2}}}} \right)\)
e) \(\dfrac{{2x + 10}}{{{x^3} - 64}}:\dfrac{{{{\left( {x + 5} \right)}^2}}}{{2x - 8}}\)
f) \(\dfrac{1}{{x + y}}\left( {\dfrac{{x + y}}{{xy}} - x - y} \right) - \dfrac{1}{{{x^2}}}:\dfrac{y}{x}\)
\(a,\dfrac{8y}{3x^2}.\dfrac{9x^2}{4y^2}=\dfrac{72x^2y}{12x^2y^2}=\dfrac{6}{y}\\b,\dfrac{3x+x^2}{x^2+x+1}.\dfrac{3x^3-3}{x+3}=\dfrac{x\left(x+3\right)3\left(x-1\right)\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left(x+3\right)}=3x\left(x-1\right)=3x^2-3x \)
\(c,\dfrac{2x^2+4}{x-3}.\dfrac{3x+1}{x-1}.\dfrac{6-2x}{x^2+2}=\dfrac{2\left(x^2+2\right)\left(3x+1\right)2\left(3-x\right)}{\left(x-3\right)\left(x-1\right)\left(x^2+2\right)}=\dfrac{-4\left(3x+1\right)}{x-1}=\dfrac{-12x-4}{x-1}\)
\(d,\dfrac{2x^2}{3y^3}:\left(-\dfrac{4x^3}{21y^2}\right)=\dfrac{-2x^2.21y^2}{3y^3.4x^3}=\dfrac{-42x^2y^2}{12x^3y^3}=\dfrac{-7}{2xy}\)
\(e,\dfrac{2x+10}{x^3-64}:\dfrac{\left(x+5\right)^2}{2x-8}=\dfrac{2\left(x+5\right)}{\left(x-4\right)\left(x^2+4x+16\right)}.\dfrac{2\left(x-4\right)}{\left(x+5\right)^2}=\dfrac{4}{\left(x+5\right)\left(x^2+4x+16\right)}=\dfrac{4}{x^3+9x^2+16x+80}\)
\(f,\dfrac{1}{x+y}\left(\dfrac{x+y}{xy}-x-y\right)-\dfrac{1}{x^2}:\dfrac{y}{x}=\dfrac{1}{x+y}\left(\dfrac{\left(x+y\right)\left(1-xy\right)}{xy}\right)-\dfrac{x}{x^2y}=\dfrac{1-xy}{xy}-\dfrac{x}{x^2y}=\dfrac{-x^2y}{x^2y}=-1\)
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\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\)\(\cdot\dfrac{12}{3^2}\cdot\dfrac{20}{4^2}\cdot\cdot\cdot\cdot\dfrac{110}{10^2}\)\(\cdot\left(x-2\right)=-20\left(x+1\right)+60\)
giải phương trình trên
\(\dfrac{1.2}{1.1}.\dfrac{2.3}{2.2}.\dfrac{3.4}{3.3}.\dfrac{4.5}{4.4}...\dfrac{10.11}{10.10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow\dfrac{2.3.4...11}{1.2.3...10}\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11\left(x-2\right)=-20x+40\)
\(\Leftrightarrow11x-22=-20x+40\)
\(\Leftrightarrow31x=62\)
\(\Rightarrow x=2\)
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\(=>\dfrac{2\cdot1}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\dfrac{3\cdot4}{3\cdot3}\cdot......\cdot\dfrac{10\cdot11}{10\cdot10}\cdot\left(x-2\right)=-20\left(x+1\right)+60\)=>11*(x-2)=-20*(x+1)+60
=>11x-22=-20x-20+60
=>31x=62
=>x=2
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Tìm x
a) 4/5 ( x - 1/3) - \(3\dfrac{1}{2}\)= 50%
b) \(8\dfrac{3}{5}\cdot x-2\dfrac{1}{5}\cdot x=16\%\)
a) \(\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)-3\dfrac{1}{2}=50\%\)
\(\dfrac{4}{5}x-\dfrac{4}{15}-\dfrac{7}{2}=\dfrac{1}{2}\)
\(\dfrac{4}{5}x=\dfrac{1}{2}+\dfrac{7}{2}+\dfrac{4}{15}\)
\(\dfrac{4}{5}x=\dfrac{64}{15}\)
\(x=\dfrac{64}{15}:\dfrac{4}{5}\\ x=\dfrac{16}{3}\)
b) \(8\dfrac{3}{5}.x-2\dfrac{1}{5}.x=16\%\)
\(\dfrac{43}{5}x-\dfrac{11}{5}x=\dfrac{4}{25}\)
\(\dfrac{32}{5}x=\dfrac{4}{25}\)
\(x=\dfrac{4}{25}:\dfrac{32}{5}\)
\(x=\dfrac{1}{40}\)
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a \(\dfrac{25x^3y}{7z}\cdot\dfrac{28z}{15x^2y^5}\)
b \(\dfrac{x^2+3x+9}{2x+10}\cdot\dfrac{x+5}{x^3+27}\)
c \(\dfrac{3x-6}{x-1}\cdot\dfrac{1-x^3}{10-5x}\)
d \(\dfrac{3x-2}{x^2+1}\cdot\dfrac{x-1-x^2}{4-9x^2}\)
a) \(\dfrac{25x^3y}{7z}.\dfrac{28z}{15x^2y^5}\)
\(=\dfrac{25x^3y.28z}{7z.15x^2y^5}\)
\(=\dfrac{700x^3yz}{105x^2y^5z}\)
\(=\dfrac{20x}{3y^4}\)
b) \(\dfrac{x^2+3x+9}{2x+10}.\dfrac{x+5}{x^3+27}\)
\(=\dfrac{\left(x^2+3x+9\right)\left(x+5\right)}{\left(2x+10\right)\left(x^3+27\right)}\)
\(=\dfrac{x^3+3x^2+9x+5x^2+15x+45}{2x^4+54x+10x^3+270}\)
\(=\dfrac{x^3+8x^2+24x+45}{2x^4+10x^3+54x+270}\)
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Xem thêm câu trả lời
Tìm $x$, biết :
a) $\left(\dfrac{1}{2}+1,5\right) \cdot x=\dfrac{1}{5}$
b) $\left(-1 \dfrac{3}{5}+x\right): \dfrac{12}{13}=2 \dfrac{1}{6}$
c) $\left(x: 2 \dfrac{1}{3}\right) \cdot \dfrac{1}{7}=\dfrac{-3}{8}$
d) $\dfrac{-4}{7} \cdot x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1 \dfrac{2}{3}\right)$
\(a)\left(\dfrac{1}{2}+1,5\right)x=\dfrac{1}{5}\)
\(\Rightarrow2x=\dfrac{1}{5}\)
\(\Rightarrow x=\dfrac{1}{10}\)
\(b)\left(-1\dfrac{3}{5}+x\right):\dfrac{12}{13}=2\dfrac{1}{6}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=\dfrac{13}{6}.\dfrac{12}{13}\)
\(\Leftrightarrow-\dfrac{8}{5}+x=2\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(c)\left(x:2\dfrac{1}{3}\right).\dfrac{1}{7}=-\dfrac{3}{8}\)
\(\Leftrightarrow x:\dfrac{7}{3}=-\dfrac{3}{8}:\dfrac{1}{7}\)
\(\Leftrightarrow x=-\dfrac{21}{8}.\dfrac{7}{3}\)
\(\Leftrightarrow x=-\dfrac{49}{8}\)
\(d)-\dfrac{4}{7}x+\dfrac{7}{5}=\dfrac{1}{8}:\left(-1\dfrac{2}{3}\right)\)
\(\Leftrightarrow-\dfrac{4}{7}x+\dfrac{7}{5}=-\dfrac{3}{40}\)
\(\Leftrightarrow-\dfrac{4}{7}x=-\dfrac{59}{40}\)
\(\Leftrightarrow x=\dfrac{413}{160}\)
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Xem thêm câu trả lời
Câu 1:
a)dfrac{3}{x-1}dfrac{4}{y-2}dfrac{5}{z-3};x+y+z18
b)dfrac{3}{x-1}dfrac{4}{y-2}dfrac{5}{z-3};xcdot ycdot z192
c)2cdot x3cdot y;5cdot y3cdot z;3cdot x+3cdot y-7cdot z35
Câu 2:Tìm 3 số biết tổng các bình phương của chúng bằng 481.Số thứ 2 bằng dfrac{4}{3}số thứ nhất và bằng dfrac{3}{4}số thứ 3
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Câu 1:
a)\(\dfrac{3}{x-1}=\dfrac{4}{y-2}=\dfrac{5}{z-3};x+y+z=18\)
b)\(\dfrac{3}{x-1}=\dfrac{4}{y-2}=\dfrac{5}{z-3};x\cdot y\cdot z=192\)
c)\(2\cdot x=3\cdot y;5\cdot y=3\cdot z;3\cdot x+3\cdot y-7\cdot z=35\)
Câu 2:Tìm 3 số biết tổng các bình phương của chúng bằng 481.Số thứ 2 bằng \(\dfrac{4}{3}\)số thứ nhất và bằng \(\dfrac{3}{4}\)số thứ 3
Câu 1:
c: 2x=3y
nên x/3=y/2
=>x/9=y/6
5y=3z
nên y/3=z/5
=>y/6=z/10
=>x/9=y/6=z/10
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{10}=\dfrac{3x+3y-7z}{3\cdot9+3\cdot6-7\cdot10}=\dfrac{35}{-25}=-\dfrac{7}{5}\)
Do đó: x=-63/5; y=-42/5; z=-14
Bài 2:
Gọi ba số lần lượt là a,b,c
Theo đề, ta có: 4/3a=b=3/4c
\(\Leftrightarrow\dfrac{a}{\dfrac{3}{4}}=\dfrac{b}{1}=\dfrac{c}{\dfrac{4}{3}}\)
\(\Leftrightarrow\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}\)
Đặt \(\dfrac{a}{9}=\dfrac{b}{12}=\dfrac{c}{16}=k\)
=>a=9k; b=12k; c=16k
Theo đề, ta có: \(a^2+b^2+c^2=481\)
\(\Leftrightarrow81k^2+144k^2+256k^2=481\)
=>k2=1
Trường hợp 1: k=1
=>a=9; b=12; c=16
Trường hợp 2: k=-1
=>a=-9; b=-12; c=-16
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