(x+1) + ( x+2) + (x+3) + (x+4) + ( x+5)= 65
Giải các phương trình sau:
a \(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
b \(\sqrt[3]{\left(65+x\right)^2}+4\sqrt[3]{\left(65-x\right)^2}=5\sqrt[3]{65^2-x^2}\)
c \(\sqrt[3]{x+1}+\sqrt[3]{x+2}=1+\sqrt[3]{x^2+3x+2}\)
d \(\sqrt[3]{x-2}+\sqrt[3]{x+3}=\sqrt[3]{2x+1}\)
e \(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=\sqrt[3]{3x+1}\)
a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
c.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x+1}=b\end{matrix}\right.\)
\(\Rightarrow a+b=1+ab\)
\(\Leftrightarrow\left(a-1\right)\left(b-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=1\\b^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=1\\x+1=1\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải phương trình
a, \(\sqrt{x-1+4\sqrt{x-5}}+\sqrt{11+x+8\sqrt{x-5}}=0\)
b, \(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2+\sqrt{2x-5}}=\sqrt{8}\)
c. \(\sqrt[3]{\left(65+x\right)^2}+4\sqrt[3]{\left(65-x\right)^2}=5\sqrt[3]{65^2-x^2}\)
d, \(\sqrt{\dfrac{x^2+x+1}{x}}+\sqrt{\dfrac{x}{x^2+x+1}}=\dfrac{7}{4}\)
b, ĐKXĐ: \(x\ge\frac{5}{2}\)
\(pt\Leftrightarrow\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}+1\right)^2}=4\)
\(\Leftrightarrow\sqrt{2x-5}=3\)
\(\Leftrightarrow x=7\left(tm\right)\)
a, ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5+8\sqrt{x-5}+16}=0\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-5}+2\right)^2}+\sqrt{\left(\sqrt{x-5}+4\right)^2}=0\)
\(\Leftrightarrow2\sqrt{x-5}+6=0\)
\(\Leftrightarrow\sqrt{x-5}=-3\)
Phương trình vô nghiệm
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 65
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 65
( x + x + x + x + x ) + ( 1 + 2 + 3 + 4 = 5 ) = 65
x * 5 +15 = 65
x * 5 = 65 - 15
x * 5 = 50
x = 50 : 5
x = 10
( X + 1 ) + ( X + 2 ) + ( X + 3 ) + ( X + 4 ) + ( X + 5 ) = 65
X x 5 + ( 1 + 2 + 3 + 4 + 5 ) = 65
X x 5 + 15 = 65
X x 5 = 65 - 15
X x 5 = 50
X = 50 : 5
X = 10
Tìm x: (x+1)+(x+2)+(x+3)+(x+4)+(x+5)=65
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)+\left(x+5\right)=65\)
<=> \(5x+15=65\)
<=> \(5x=50\)
<=> \(x=10\)
Vậy...
p/s: chúc bạn học tốt
Tìm x biết [x+1]+[x+2]+[x+3]+[x+4]+[x+5]=65
\(5x+\frac{5.6}{2}=65\Rightarrow x=\frac{65-15}{5}=10\)
1. Tính nhanh : A = 1/(1+2 ) + 1/(1+2+3) + 1/(1+2+3+4) + 1/(1+2+3+4+5) + .... + 1/(1+2+3+...+10)
2. So sánh A và B : A = ( 11 x 13 x 15 + 33 x 39 x 45 + 55 x 65 x 75 + 99 x 117 x 135 ) : ( 11 x 13x 17 + 39 x 45 x 57 + 65 x 75 x 85 + 117 x 135 x 153 ) B = 1111 : 1717
Bài 1:
\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)
\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(A=\frac{9}{11}\)
Bài 2 :
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)
\(=\frac{11}{17}\)
\(=\frac{1111}{1717}=B\)
Vậy \(A=B\)
tim x
\dfrac{5}{6}x-\dfrac{3}{4}=\dfrac{-1}{4}+\dfrac{2}{3}65x−43=4−1+32
-1\dfrac{1}{2}-\dfrac{2}{3}x=\dfrac{5}{6}-\left(\dfrac{-2}{5}\right)−121−32x=65−(5−2)
\left(\dfrac{4}{5}:x+1,5\right):\dfrac{2}{3}=-1,5(54:x+1,5):32=−1,5
\dfrac{4}{3}x-\dfrac{2}{3}=\dfrac{1}{4}-x34x−32=41−x
giup minh nhe minh dang can gap
Tim x: 6^2-(x+3)=45
125-5.(3x-1)=5^5.5^3
4^x+1+4^0=65
70-5.(x-3)=5.3^2
\(6^2-\left(x+3\right)=45\)
\(36-\left(x+3\right)=45\)
\(x+3=35-45\)
\(x+3=-10\)
\(x=-13\)
\(6^2-\left(x+3\right)=45\)
\(\Rightarrow36-\left(x+3\right)=45\)
\(\Rightarrow x+3=36-45\)
\(\Rightarrow x+3=-9\)
\(\Rightarrow x=-9-3=-12\)
\(125-5\left(3x-1\right)=5^5.5^3\)
\(\Rightarrow5^3-5\left(3x-1\right)=5^5.5^3\)
\(\Rightarrow5^3-15x+5=5^5.5^3\)
\(\Rightarrow5\left(5^2-3x+1\right)=5^5.5^3\)
\(\Rightarrow5^2-3x+1=5^5.5^3:5\)
\(\Rightarrow25-3x+1=5^7\)
Từ đây làm nốt nhé
\(4^x+1+4^0=65\)
\(\Rightarrow4^{x+1}+1=65\)
\(\Rightarrow4^{x+1}=64\)
\(\Rightarrow4^{x+1}=4^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
\(70-5\left(x-3\right)=5.3^2\)
\(\Rightarrow70-5\left(x-3\right)=5.9=45\)
\(\Rightarrow5\left(x-3\right)=70-45=25\)
\(\Rightarrow x-3=25:5=5\)
\(\Rightarrow x=8\)
Chúc em học tốt hơn nhé!!
1. Tính nhanh :
A = 1/(1+2 ) + 1/(1+2+3) + 1/(1+2+3+4) + 1/(1+2+3+4+5) + .... + 1/(1+2+3+...+10)
2. So sánh A và B :
A = ( 11 x 13 x 15 + 33 x 39 x 45 + 55 x 65 x 75 + 99 x 117 x 135 ) : ( 11 x 13x 17 + 39 x 45 x 57 + 65 x 75 x 85 + 117 x 135 x 153 )
B = 1111 : 1717
1)\(A=\frac{1}{3}+\frac{1}{6}+...+\frac{1}{55}=2\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{110}\right)=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\right)=2.\left(\frac{1}{2}-\frac{1}{11}\right)=\frac{9}{11}\)
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
=> \(A
giải giùm mình bài này vớC= (67/111+2/33-15/117)x(1/3-1/4-1/12)