\(1.\text{Tìm x:}\)
\(-7\left(3x-5\right)+2\left(7x-14\right)=28\)
\(\text{2.Tính}\)
\(A=6^2+6^4+6^6+....+6^{2010}\)
\(\text{P/s: Đăng_hộ_e_gái}\)
\(\text{https://olm.vn/thanhvien/camvancamly}\)
Những câu hỏi liên quan
A=\(\frac{5x\left(2^2\text{x}3^2\right)^9\text{x}\left(2^2\right)^6-2\text{x}\left(2^2\text{x}3\right)^{14}\text{x}3^4}{\text{ }5\text{x}2^{28}\text{x}3^{18}-7\text{x}2^{29}\text{x}3^{18}}\)
\(\frac{5.2^{18}.3^{18}.2^{12}-2.2^{28}.3^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{5.2^{30}.3^{18}-2^{29}.3^{18}}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}=\frac{2^{29}.3^{18}\left(5.2-1\right)}{2^{28}.3^{18}\left(5-7.2\right)}\)
\(\frac{2^{29}.3^{18}.9}{2^{28}.3^{18}.-9}=\frac{2.9}{-9}=-2\)
Tính nhanh
a) \(\frac{6}{7}:\left(\frac{1}{2}\text{X}\frac{3}{4}\right)-\frac{5}{8}\)
b) 34-2:\(\left(\frac{3}{5}-\frac{1}{2}\right)\)
c) \(\left(4\frac{1}{2}+\frac{1}{2}:5\frac{1}{2}\right)\text{X}\left(3\frac{5}{6}+2\frac{1}{6}\text{x}6\right)\)
Tính:1. frac{x^2}{x^2-x}-frac{x^2}{x+1}-frac{2text{x}}{x^2-1}2. frac{4x^2-3x+5}{x^3-1}-frac{1-2text{x}}{x^2+x+1}-frac{6}{x-1}3. frac{5}{2text{x}^2+6text{x}}-frac{4-3text{x}^2}{x^2-9}-34. frac{7}{8x^2-18}+frac{1}{2text{x}^2+3text{x}}-frac{1}{4text{x}-6}5. frac{1}{xleft(x+1right)}+frac{1}{left(x+1right)left(x+2right)}+...+frac{1}{left(x+9right)left(x+10right)}
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Tính:
1. \(\frac{x^2}{x^2-x}-\frac{x^2}{x+1}-\frac{2\text{x}}{x^2-1}\)
2. \(\frac{4x^2-3x+5}{x^3-1}-\frac{1-2\text{x}}{x^2+x+1}-\frac{6}{x-1}\)
3. \(\frac{5}{2\text{x}^2+6\text{x}}-\frac{4-3\text{x}^2}{x^2-9}-3\)
4. \(\frac{7}{8x^2-18}+\frac{1}{2\text{x}^2+3\text{x}}-\frac{1}{4\text{x}-6}\)
5. \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+...+\frac{1}{\left(x+9\right)\left(x+10\right)}\)
Tìm x:
\(x-\text{[}\left(\frac{3x+3}{4}\right):2\text{]}=\text{[}\left(\frac{9+x}{6}\right):2\text{]} \)
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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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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\(\left[6+7x\right]^2\left[3x+4\right]\left[x+1\right]=1\)
\(\left[x-1\right]^4\left[x+3\right]^{\text{4 }}=82\)
Tính bằng cách thuận tiện nhất:
\(\left(1-\dfrac{1}{2}\right)\text{× }\left(1-\dfrac{1}{3}\right)\text{ × }\left(1-\dfrac{1}{4}\right)\text{ × }\left(1-\dfrac{1}{5}\right)\text{}\text{}\text{× }\left(1-\dfrac{1}{6}\right)\)
Lời giải:
Gọi tích trên là $A$. Ta có:
$A=\frac{1}{2}\times \frac{2}{3}\times \frac{3}{4}\times \frac{4}{5}\times \frac{5}{6}$
$=\frac{1\times 2\times 3\times 4\times 5}{2\times 3\times 4\times 5\times 6}=\frac{1}{6}$
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Chó biểu thức : text{A}frac{left(2-1frac{2}{5}right)^2+left|1-frac{12}{5}right|-left(-frac{2}{5}right)^2}{frac{1}{2}-frac{3}{7}:left(-2frac{1}{7}right)+left|frac{1}{5}-frac{1}{2}right|-frac{2}{5}}.text{B}frac{6:frac{3}{5}+1frac{1}{6}cdotfrac{6}{7}+left(frac{1}{3}-frac{1}{4}-frac{1}{12}right)}{frac{4}{11}-frac{2}{22}+5frac{2}{11}-left(frac{6}{-11}right)}.Tính : frac{text{A}}{text{B}}.(Trích đề thi HKI 7_Cô giáo dạy toán Phạm Thị Hồng Duyên-THCS Nghiêm Xuyên--Huyện Thường Tín--Tp. Hà Nội)
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Chó biểu thức :
\(\text{A}=\frac{\left(2-1\frac{2}{5}\right)^2+\left|1-\frac{12}{5}\right|-\left(-\frac{2}{5}\right)^2}{\frac{1}{2}-\frac{3}{7}:\left(-2\frac{1}{7}\right)+\left|\frac{1}{5}-\frac{1}{2}\right|-\frac{2}{5}}\).
\(\text{B}=\frac{6:\frac{3}{5}+1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)}{\frac{4}{11}-\frac{2}{22}+5\frac{2}{11}-\left(\frac{6}{-11}\right)}\).
Tính : \(\frac{\text{A}}{\text{B}}\).
(Trích đề thi HKI 7_Cô giáo dạy toán Phạm Thị Hồng Duyên-THCS Nghiêm Xuyên--Huyện Thường Tín--Tp. Hà Nội)