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\(\sqrt{\left(1-\sqrt{50}\right)^2}=\sqrt{50}-1\approx6,07>6\)
\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2}>6\)
Ta có:\(\sqrt{\left(1-\sqrt{50}\right)^2}=|1-\sqrt{50}|=\sqrt{50}-1>\sqrt{49}-1=7-1=6\)
\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2>6}\)
Tính:
\(3.\left(x-2\right)-4.\left(2x+1\right)-5.\left(2x+3\right)=50\)
\(\Rightarrow3x-6-\left(8x+4\right)-\left(10x+15\right)=50\)
\(\Rightarrow3x-6-8x-4-10x-15=50\)
\(\Rightarrow-15x-25=50\)
\(\Rightarrow-15x=50+25\)
\(\Rightarrow-15x=75\)
\(\Rightarrow x=75:\left(-15\right)\)
\(\Rightarrow x=-5.\)
Vậy \(x=-5.\)
Chúc bạn học tốt!
\(3x^2-3xy-y-5x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)-y-5x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)+x-y-6x=-20\)
\(\Rightarrow\)\(3x\left(x-y\right)+\left(x-y\right)-6x=-20\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-6x=-20\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-6x-2=-22\)
\(\Rightarrow\)\(\left(x-y\right)\left(3x+1\right)-\left(6x+2\right)=-22\)
\(\Rightarrow\left(x-y\right)\left(3x+1\right)-2\left(3x+1\right)=-22\)
\(\Rightarrow\left(3x+1\right)\left(x-y-2\right)=-22\)
Ta có bảng sau:
| \(3x+1\) | \(-1\) | \(1\) | \(-22\) | \(22\) |
| \(x\) | \(x\notin Z\) | \(0\) | \(x\notin Z\) | \(7\) |
| \(x-y-2\) | \(-22\) | \(-1\) | ||
| \(y\) | \(-20\) | \(6\) |
Vậy ta có 2 bộ (x,y) là (0;-20) và (7;6)
Chúc bạn học tốt!
Ta có: \(1=\sqrt{1}< \sqrt{50}\Rightarrow1-\sqrt{50}< 0\)
\(\Rightarrow\sqrt{\left(1-\sqrt{50}\right)^2}=\sqrt{50}-1>\sqrt{49}-1=7-1=6\)
Vậy \(\sqrt{\left(1-\sqrt{50}\right)^2}>6\)
Bài này là bài của lớp 9 nha!! có chỗ nào ko hiểu ib
\(a,A=\sqrt{18}+\sqrt{50}-\frac{1}{2}\sqrt{98}.\)
\(=3\sqrt{2}+5\sqrt{2}-\frac{7}{2}\sqrt{2}\)
\(=\sqrt{2}\left(3+5-\frac{7}{2}\right)\)
\(=\frac{9}{2}\sqrt{2}\)
\(b,B=\left(2\sqrt{3}+7\right)\left(2\sqrt{3}-7\right)\)
\(=2^2\sqrt{3^2}-7^2\)
\(=12-49=-37\)
a )
\(A=\sqrt{18}+\sqrt{50}-\frac{1}{2}\sqrt{98}\)
\(A=3\sqrt{2}+5\sqrt{2}-\frac{7}{2}\sqrt{2}\)
\(A=(3+5-\frac{7}{2})\sqrt{2}\)
\(A=\frac{9}{2}\sqrt{2}=\frac{9\sqrt{2}}{2}\)
b)
\(B=\left(2\sqrt{3}+7\right)\left(2\sqrt{3}-7\right)=\left(2\sqrt{3}\right)^2-7^2=12-49=-37\)
\(\sqrt{\frac{\left(-5\right)^2}{7}}=\frac{\sqrt{\left(-5\right)^2}}{\sqrt{7}}=\frac{|5|}{\sqrt{7}}=\frac{5\sqrt{7}}{7}\)
\(\frac{-\sqrt{\left(-5\right)^2}}{-\sqrt{49}}=\frac{\sqrt{\left(-5\right)^2}}{\sqrt{49}}=\frac{|5|}{|7|}=\frac{5}{7}\)
\(\frac{5\sqrt{7}}{7}>\frac{5}{7}\leftrightarrow\sqrt{\frac{\left(-5\right)^2}{7}}>\frac{-\sqrt{\left(-5\right)^2}}{-\sqrt{49}}\)
a, \(7+\sqrt{5}\) ta co \(\sqrt{5}>\sqrt{4}\)(1)
\(\sqrt{48}+2\) \(\sqrt{48}<\sqrt{49}\)(2)
\(7+\sqrt{4}=7+2=9\)(3)
\(\sqrt{49}+2=7+2=9\)(4)
tu (1);(2);(3);(3) = > lam not di
b,\(1-\sqrt{50}\) cung so sanh \(\sqrt{50}voi\sqrt{49}\) tu lam not nha
k dung minh nha