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a)5x+5x+2=650
\(\Rightarrow5^x\left(1+5^2\right)=650\)
\(\Rightarrow5^x\cdot26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
b)\(3^{x-1}+5\cdot3^{x-1}=162\)
\(\Rightarrow3^{x-1}\cdot\left(1+5\right)=162\)
\(\Rightarrow3^{x-1}\cdot6=162\)
\(\Rightarrow3^{x-1}=27\)
\(\Rightarrow3^{x-1}=3^3\)
\(\Rightarrow x-1=3\)
\(\Rightarrow x=4\)
a: 2x(x+1)-135=-200
=>2(x^2+x)=-65
=>2x^2+2x+65=0
=>x^2+x+32,5=0
=>x^2+x+0,25+32,25=0
=>(x+0,5)^2+32,25=0(vô lý)
b: 4x-5(x-1)+15=13
=>4x-5x+5=-2
=>5-x=-2
=>x=5+2=7
c: 2/3x-1/4=3/5-7/8
=>2/3x=3/5-7/8+1/4=24/40-35/40+10/40=-1/40
=>x=-1/40:2/3=-1/40*3/2=-3/80
d: 1/2(2x-3)+105/2=-137/2
=>1/2(2x-3)=-137/2-105/2=-242/2=-121
=>2x-3=-242
=>2x=-239
=>x=-239/2
Bài 1:
Đặt \(a=\sqrt[7]{\dfrac{3}{5}};b=\sqrt[7]{\dfrac{5}{3}}\Rightarrow\left\{{}\begin{matrix}a+b=x\\ab=1\end{matrix}\right.\)
Ta có \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\)
\(\Rightarrow a^3+b^3=x\left(x^2-3\right)=x^3-3x\)
Ta có \(a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=\left[\left(a+b\right)^2-2ab\right]^2-2\left(ab\right)^2\)
\(\Rightarrow a^4+b^4=\left(x^2-2\right)^2-2=x^4-4x^2+2\)
\(\Rightarrow\left(a^3+b^3\right)\left(a^4+b^4\right)=\left(x^3-3x\right)\left(x^4-4x^2+2\right)\\ =x^7-3x^5-4x^5+12x^3+2x^3-6x\\ =x^7-7x^5+14x^3-6x\)
Lại có \(\left(a^4+b^4\right)\left(a^3+b^3\right)=a^7+b^7+\left(ab\right)^3\left(a+b\right)=\dfrac{3}{5}+\dfrac{5}{3}+x=\dfrac{34}{15}+x\)
\(\Rightarrow x^7-7x^5+14x^3-6x=\dfrac{34}{15}+x\\ \Rightarrow15x^7-105x^5+210x^3-105x-34=0\left(1\right)\)
Vậy (1) nhận \(x=\sqrt[7]{\dfrac{3}{5}}+\sqrt[7]{\dfrac{5}{3}}\) làm nghiệm
Bài 2 đa thức bậc 2 chia đa thức bậc 2 dư đa thức bậc 1 ??
Bài 1:
a) \(3\left(x+5\right)=x-7\)
\(\Leftrightarrow3x+15=x-7\)
\(\Leftrightarrow3x+15-x=-7\)
\(\Leftrightarrow2x+15=-7\)
\(\Leftrightarrow2x=-22\)
\(\Leftrightarrow x=-11\)
Vậy \(x=-11\)
Bài 2:
\(\left|x+2\right|-14=-9\)
\(\Leftrightarrow\left|x+2\right|=5\)
Chia 2 trường hợp:
\(\Leftrightarrow\orbr{\begin{cases}x+2=5\\x+2=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-7\end{cases}}}\)
Vậy \(x\in\left\{3;-7\right\}\)
Hơi vội, sai thì thôi nhé!
a. -3
b. -71
c. -40
d. 358
e. ?i
f. 10
mình chỉ cho đáp án được thôi vì mình đang gấp nên ko thể trình bày được, câu e mình ko hiểu bạn ghi gì hết. bạn gửi lại câu e đi.
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a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
a) x - 7 = 5. 0 => x - 7 = 0 =>x = 7.
b) x: 3 = 47 +13 => x: 3 = 60 => x = 60.3 => x = 180.
c) x : 7 - 7 = 0 hoặc x : 12 - 12 = 0. Do đó x = 49 hoặc x = 144.
d) x : 2 = 150 - 135 => x: 2 = 15 => x = 15.2 => x = 30.
e) 100: x = 140 -120 => 100: x = 20 => x = 100:20 => x = 5.
g) x : 5 = 300 - 273 => x : 5 = 27 =>x = 27.5 => x = 135
Bài 1:
a) \(\dfrac{9}{20}-\dfrac{8}{15}\times\dfrac{5}{12}\)
\(=\dfrac{9}{20}-\dfrac{2}{9}\)
\(=\dfrac{41}{180}\)
b) \(\dfrac{2}{3}\div\dfrac{4}{5}\div\dfrac{7}{12}\)
\(=\dfrac{2}{3}\times\dfrac{5}{4}\times\dfrac{12}{7}\)
\(=\dfrac{5}{6}\times\dfrac{12}{7}\)
\(=\dfrac{10}{7}\)
c) \(\dfrac{7}{9}\times\dfrac{1}{3}+\dfrac{7}{9}\times\dfrac{2}{3}\)
\(=\dfrac{7}{9}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\)
\(=\dfrac{7}{9}\times1\)
\(=\dfrac{7}{9}\)
Bài 2:
a) \(2\times\left(x-1\right)=4026\)
\(\left(x-1\right)=4026\div2\)
\(x-1=2013\)
\(x=2014\)
Vậy: \(x=2014\)
b) \(x\times3,7+6,3\times x=320\)
\(x\times\left(3,7+6,3\right)=320\)
\(x\times10=320\)
\(x=320\div10\)
\(x=32\)
Vậy: \(x=32\)
c) \(0,25\times3< 3< 1,02\)
\(\Leftrightarrow0,75< 3< 1,02\) ( S )
=> \(0,75< 1,02< 3\)
b)
X : 5 = 135 ( dư 3 )
X = 135 x 5
X = 675
X = 675 + 3
X = 678
b) 5 x X = 4680
X = 4680 : 5
X = 930
c) ( x : 7 ) x 3 = 162
X x 3 = 162 x 7
X x 3 = 1134
X = 1134 : 3
X = 378