1/ Rút gọn biểu thức sau: B= \(\dfrac{35^3+5.35^2-5^3.7}{10.70^2+10^2.70-10^3}\)
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\(B=\frac{35^3+5.35^2-5^3.7}{10.70^2+10^2.70-10^3}=\frac{5^3.7^3+5^3.7^2-5^3.7}{10^3.7^2+10^3.7-10^3}=\frac{5^3.7.\left(7^2+7-1\right)}{10^3.\left(7^2+7-1\right)}.\)
=> \(B=\frac{5^3.7}{10^3}=\frac{5^3.7}{2^3.5^3}=\frac{7}{2^3}=\frac{7}{8}\)
1) ta có x.y=-30=>y=\(-\frac{30}{x}\)
z-x=-12=> z=-12-x
nên y.z=\(-\frac{30}{x}.\left(-12-x\right)=42\)
\(=\frac{360}{x}-\frac{30x}{x}=42\)
\(=\frac{360-30x}{x}=42\)
\(=>360-30x=42x\)
\(=360-30x-42x=0\)
\(=360-72x=0\)
\(< =>72x=360\)
\(x=5\)=> \(y=-6\); \(z=-7\)
1.Ta có A= 710 +79 - 78
A= 78 .(72 +7 -1)
A=78 .55
=> A chia hết cho 11( vì có thừa số 55 chia hết cho 11)
a) ta có : \(z-x=-12\Leftrightarrow z=x-12\)
\(\Rightarrow yz=42\Leftrightarrow y\left(x-12\right)=42\Leftrightarrow xy-12y=42\)
\(\Leftrightarrow-30-12y=42\Leftrightarrow12y=-30-42=-72\Leftrightarrow y=\dfrac{-72}{12}=-6\)
ta có : \(y=-6\Rightarrow xy=-30\Leftrightarrow x.-6=-30\Leftrightarrow x=\dfrac{-30}{-6}=5\)
ta có : \(x=5\Rightarrow z=5-12=-7\)
vậy \(x=5;y=-6;z=-7\)
b) ta có :\(A=7^{10}+7^9-7^8=7^8.\left(7^2+7-1\right)=7^8.55=7^8.5.11⋮11\)
\(\Leftrightarrow7^8.5.11\) chia hết cho \(11\) \(\Leftrightarrow\) A chia hết cho 11
vậy A chia hết cho 11 (đpcm)
a)xy=30 ;yz=42=>\(y=\dfrac{30}{x}\);\(y=\dfrac{42}{z}\)
Do đó \(\dfrac{30}{x}=\dfrac{42}{z}\)
Áp dụng t/c của dãy tỉ số bằng nhau,tac có:
\(\dfrac{30}{x}=\dfrac{42}{z}\)=\(\dfrac{42-30}{z-x}\)=\(\dfrac{12}{-12}=-1\)
=>x=-30;z=-42
Do đó y=\(\dfrac{30}{x}=\dfrac{30}{-30}=-1\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
\(=\dfrac{5-3\sqrt{5}+10+6\sqrt{5}}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}-\dfrac{2\sqrt{10}+2}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{15+3\sqrt{5}}{5-9}-\left(2\sqrt{10}+2\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =-2\sqrt{30}-4\sqrt{5}-2\sqrt{3}-2\sqrt{2}-\dfrac{15+3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-16\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15-3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-19\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15}{4}\)
b: Ta có: \(\dfrac{4}{\sqrt{3}+1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{6}{3-\sqrt{3}}\)
\(=2\sqrt{3}-2+\sqrt{3}+1-3-\sqrt{3}\)
\(=2\sqrt{3}-4\)
\(B=\dfrac{35^3+5\cdot35^2-5^3\cdot7}{10\cdot70^2+10^2\cdot70-10^3}=\dfrac{\left(5\cdot7\right)^3+5\cdot\left(5\cdot7\right)^2-5^3\cdot7}{2\cdot5\cdot\left(2\cdot5\cdot7\right)^2+\left(2\cdot5\right)^2\cdot2\cdot5\cdot7-\left(2\cdot5\right)^3}=\dfrac{5^3\cdot7^3+5\cdot5^2\cdot7^2-5^3\cdot7}{2\cdot5\cdot2^2\cdot5^2\cdot7^2+2^2\cdot5^2\cdot2\cdot5\cdot7-2^3\cdot5^3}=\dfrac{5^3\cdot7^3+5^3\cdot7^2-5^3\cdot7}{2^3\cdot5^3\cdot7^2+2^3\cdot5^3\cdot7-2^3\cdot5^3}=\dfrac{5^3\left(7^3+7^2-7\right)}{2^3\cdot5^3\left(7^2+7-1\right)}=\dfrac{343+49-7}{8\cdot\left(49+7-1\right)}=\dfrac{385}{8\cdot55}=\dfrac{385}{440}=\dfrac{7}{8}\)
Vậy \(B=\dfrac{7}{8}\)