\(TH\text{ỰC}-HI\text{ỆN}-PH\text{ÉP}-T\text{ÍNH}\)
\(6⋮\left(x-1\right)\)
Những câu hỏi liên quan
\(th\text{ực}-hi\text{ện}-ph\text{ép}-t\text{ính}\)
\(\left(2002\right)-\left(57-2002\right)\)
Khỏi cần ngoặc nha Songoku Sky Fc11
2002 - 57 - 2002
= 2002 - 2002 - 57 = -57
k cho bạn Trần Nhật Quỳnh nha bạn ấy làm đúng rồi
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2002 - 57 - 2002
= 2002 - 2002 - 57
= 0 - 57
= -57
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\(\left(2002\right)-\left(57-2002\right)\)
\(=2002-2002-57\)
\(=\)\(0-57\)
\(=-57\)
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Xem thêm câu trả lời
\(TH\text{ỰC}\)\(HI\text{ỆN}\)\(PH\text{ÉP}\)\(T\text{ÍNH}\)
\(117::\left[2×\left(4^2-9\right)+3^2×\left(15-10\right)\right]\)
CÁC BẠN GIẢI DÙM MÌNH NHA
THực hiện phép tính:
117 : [ 2 x ( 4\(^2\)-9 ) + 3\(^2\). ( 15 - 10 ) ]
= 117 : [ 2 . ( 16 - 9 ) + 9 . 5]
= 117 : 2 . 8 + 45
= 58,5 .8 + 45
= 292,5 + 45
= 337,5
Tk và kb hộ mình nha m.n! thanks
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Xem thêm câu trả lời
F (x) = 3/2X2 \(choh\text{àm}s\text{ố}\text{đ}\text{ồ}th\text{ị}f\left(x\right)=-\frac{3}{2}x^2+5.t\text{ính f(-4)}\)
Bài 1: Tính
Asqrt{5-2text{√}6}+sqrt{5+2text{√}6}
B left(sqrt{10}+sqrt{6}right)sqrt{8-2text{√}15}
Csqrt{4+text{√}7}+sqrt{4-text{√}7}
Dleft(3+text{√}5right)left(text{√}10-text{√}2right)sqrt{3-text{√}5}
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a0
b, x-2sqrt{xy}+y left(xge0;yge0right)
c, sqrt{xy}+2text{√}x-3text{√}y-6left(xge0;yge0right)
Bài 3: Rút gọn
M left(frac{1}{text{√}x-1}-frac{1}{text{√}x}right)divleft(frac{text{√}x+1}{text{√}x-2}-frac{text{√}x+2}{text{√}x-1}right)
a, Rút gọn...
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Bài 1: Tính
A=\(\sqrt{5-2\text{√}6}+\sqrt{5+2\text{√}6}\)
B= \(\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\text{√}15}\)
C=\(\sqrt{4+\text{√}7}+\sqrt{4-\text{√}7}\)
D=\(\left(3+\text{√}5\right)\left(\text{√}10-\text{√}2\right)\sqrt{3-\text{√}5}\)
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a>=0
b, x-2\(\sqrt{xy}\)+y \(\left(x\ge0;y\ge0\right)\)
c, \(\sqrt{xy}+2\text{√}x-3\text{√}y-6\)\(\left(x\ge0;y\ge0\right)\)
Bài 3: Rút gọn
M= \(\left(\frac{1}{\text{√}x-1}-\frac{1}{\text{√}x}\right)\div\left(\frac{\text{√}x+1}{\text{√}x-2}-\frac{\text{√}x+2}{\text{√}x-1}\right)\)
a, Rút gọn M
b, Tính giá trị của M khi x=2
c, Tìm x để M>0
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
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Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
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Bài 3:
a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)
\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b)
Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)
c)
Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)
Kết hợp với ĐKXĐ suy ra $x>4$
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\(t\text{ính}t\text{ổng}:\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2007}\)
cho các số x,y,z khác 0 va thoả mãn :\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0.t\text{ính}gi\text{á}tr\text{ị}bi\text{ểu}th\text{ức}P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\\\frac{1}{y}+\frac{1}{z}=-\frac{1}{x}\\\frac{1}{x}+\frac{1}{z}=-\frac{1}{y}\end{cases}}\)
\(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
\(=\frac{y}{x}+\frac{z}{x}+\frac{z}{y}+\frac{x}{y}+\frac{x}{z}+\frac{y}{z}\)
\(=y\left(\frac{1}{x}+\frac{1}{z}\right)+x\left(\frac{1}{z}+\frac{1}{y}\right)+z\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=y.\frac{-1}{y}+x.\frac{-1}{x}+z.\frac{-1}{z}\)
\(=-1-1-1=-3\)
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P+3=\(\frac{y+z}{x}+1+\frac{x+z}{y}+1+\frac{x+y}{z}+1=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)
P+3=\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=0.\left(x+y+z\right)=0\)
=> P=\(-3\)
Chuc ban hoc tot
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Ta có : \(P=\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
\(\Rightarrow P+3=\frac{y+z}{x}+1+\frac{z+x}{y}+1+\frac{x+y}{z}+1\)
\(\Rightarrow P+3=\frac{x+y+z}{x}+\frac{x+y+z}{y}+\frac{x+y+z}{z}\)
\(\Rightarrow P+3=\left(x+y+z\right).\frac{1}{x}+\left(x+y+z\right).\frac{1}{y}+\left(x+y+z\right).\frac{1}{z}\)
\(\Rightarrow P+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow P+3=\left(x+y+z\right).0\)
\(\Rightarrow P+3=0\)
\(\Rightarrow P=-3\)
Vậy P = - 3
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Ttext{ìm} stext{ố}.nguytext{ê}n.dtext{ư}text{ơ}ng.nhtext{ỏ}.nhtext{ất}.thtext{ỏa}.mtext{ãn}:frac{1}{2}stext{ố}.text{đ}text{ó}.ltext{à}.stext{ố}.chtext{ính}.phtext{ư}text{ơ}ng frac{1}{3}stext{ố}.text{đ}text{ó}.ltext{à}.ltext{ập}.phtext{ư}text{ơ}ng.ctext{ủa}.1.stext{ố}.nguytext{ên} frac{1}{5}stext{ố}.text{đ}text{ó}.ltext{à}.ltext{ũy}.thtext{ừa}.5.ctext{ủa}.1.stext{ố.nguytext{ê}n}
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\(T\text{ìm}\) \(s\text{ố}.nguy\text{ê}n.d\text{ư}\text{ơ}ng.nh\text{ỏ}.nh\text{ất}.th\text{ỏa}.m\text{ãn}:\frac{1}{2}s\text{ố}.\text{đ}\text{ó}.l\text{à}.s\text{ố}.ch\text{ính}.ph\text{ư}\text{ơ}ng\) \(\frac{1}{3}s\text{ố}.\text{đ}\text{ó}.l\text{à}.l\text{ập}.ph\text{ư}\text{ơ}ng.c\text{ủa}.1.s\text{ố}.nguy\text{ên}\) \(\)
\(\frac{1}{5}s\text{ố}.\text{đ}\text{ó}.l\text{à}.l\text{ũy}.th\text{ừa}.5.c\text{ủa}.1.s\text{ố.nguy\text{ê}n}\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
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A=\(\frac{5x\left(2^2\text{x}3^2\right)^9\text{x}\left(2^2\right)^6-2\text{x}\left(2^2\text{x}3\right)^{14}\text{x}3^4}{\text{ }5\text{x}2^{28}\text{x}3^{18}-7\text{x}2^{29}\text{x}3^{18}}\)
\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-7.2\right)}\)
\(\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.-9}=\frac{2.9}{-9}=-2\)