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c.
545 - 544 = 544 x (54 - 1) = 544 x 53
Vậy 545 - 544 chia hết cho 53.
đặt \(\hept{\begin{cases}\sqrt[3]{3x-2}=a\\\sqrt{6-5x}=b\ge0\end{cases}}\) ta sẽ có hệ sau \(\hept{\begin{cases}3a+4b=10\\5a^3+3b^2=8\end{cases}}\)
rút thế \(b=\frac{10-3a}{4}\)xuống phương trình dưới ta có\
\(5a^3+3\left(\frac{10-3a}{4}\right)^2=8\) hay
\(80a^3+27a^2-180a+172=0\Leftrightarrow\left(a+2\right)\left(80a^2-133a+86\right)=0\Leftrightarrow a=-2\)
hay \(\sqrt[3]{3x-2}=-2\Leftrightarrow x=-2\) thay lại thỏa mãn
vậy phương trình có nghiệm duy nhất x=-2
4. đặt \(\sqrt[3]{x+24}=a\) và \(\sqrt{12-x}=b\)(b>=0)
==>ta có hệ pt
\(\int_{a^3+b^2=36}^{a+b=6}\)<=> \(\int_{a^3+\left(6-a\right)^2=36}^{b=6-a}\)<=> \(\int_{b=6-a}^{a^3+a^2-12a=0}\)<=> \(\int_{b=6-a}^{a\left(a^2+a-12\right)=0}\)<=>\(\int_{b=6-a}^{a\left(a+4\right)\left(a-3\right)=0}\)
đến đây bạn tự tìm a;b rufit hay vào tìm x là ok
3. \(\Leftrightarrow\sqrt[3]{2x^2}-\sqrt[3]{x+1}+\sqrt[3]{2x^2+1}-\sqrt[3]{x+2}=0\)
\(\Leftrightarrow\frac{2x^2-x-1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{2x^2-x-1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}=0\)
\(\Leftrightarrow2x^2-x-1=0\)
( do \(\frac{1}{\sqrt[3]{4x^4}+\sqrt[3]{2x^2\left(x+1\right)}+\sqrt[3]{\left(x+1\right)^2}}+\frac{1}{\sqrt[3]{\left(2x^2+1\right)^2}+\sqrt[3]{\left(2x^2+1\right)\left(x+2\right)}+\sqrt[3]{\left(x+2\right)^2}}>0\forall xTMĐK\))
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2=\frac{9}{8}\Leftrightarrow\left(x-\frac{1}{4}\right)^2=\frac{9}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{4}=\frac{3}{4}\\x-\frac{1}{4}=-\frac{3}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\frac{1}{2}\end{matrix}\right.\) ( TM )
\(\Leftrightarrow\left(\dfrac{2}{3}+\dfrac{x}{5}\right)\cdot30=80\)
=>1/5x+2/3=8/3
=>1/5x=2
hay x=10
b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{2}+2}-\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) (vì \(\sqrt{5}\ge\sqrt{2}\)
=0
c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)
\(=\sqrt{3}-1+\sqrt{3+1}\) (vì \(\sqrt{3}\ge1\))
\(=2\sqrt{3}\)
a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}-\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)
\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) (vì \(\sqrt{3}\ge\sqrt{2}\))
=0
Lời giải:
ĐK: $x\neq -1; x\neq -4$
PT \(\Leftrightarrow \frac{3}{x^2-x+1}-\frac{27}{x^2+5x+4}+\frac{11}{x^2+x+2}-\frac{27}{x^2+5x+4}=0\)
\(\Leftrightarrow \frac{3(x^2+5x+4)-27(x^2-x+1)}{(x^2-x+1)(x^2+5x+4)}+\frac{11(x^2+5x+4)-27(x^2+x+2)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{3(-8x^2+14x-5)}{(x^2-x+1)(x^2+5x+4)}+\frac{2(-8x^2+14x-5)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{-8x^2+14x-5}{x^2+5x+4}\left(\frac{3}{x^2-x+1}+\frac{2}{x^2+x+2}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ với mọi $x\neq -1; x\neq -4$
Do đó \(\frac{-8x^2+14x-5}{x^2+5x+4}=0\Rightarrow -8x^2+14x-5=0\)
\(\Rightarrow x=\frac{1}{2}\) hoặc $x=\frac{5}{4}$ (đều thỏa mãn)
Vậy........
Lời giải:
ĐK: $x\neq -1; x\neq -4$
PT \(\Leftrightarrow \frac{3}{x^2-x+1}-\frac{27}{x^2+5x+4}+\frac{11}{x^2+x+2}-\frac{27}{x^2+5x+4}=0\)
\(\Leftrightarrow \frac{3(x^2+5x+4)-27(x^2-x+1)}{(x^2-x+1)(x^2+5x+4)}+\frac{11(x^2+5x+4)-27(x^2+x+2)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{3(-8x^2+14x-5)}{(x^2-x+1)(x^2+5x+4)}+\frac{2(-8x^2+14x-5)}{(x^2+x+2)(x^2+5x+4)}=0\)
\(\Leftrightarrow \frac{-8x^2+14x-5}{x^2+5x+4}\left(\frac{3}{x^2-x+1}+\frac{2}{x^2+x+2}\right)=0\)
Dễ thấy biểu thức trong ngoặc lớn luôn lớn hơn $0$ với mọi $x\neq -1; x\neq -4$
Do đó \(\frac{-8x^2+14x-5}{x^2+5x+4}=0\Rightarrow -8x^2+14x-5=0\)
\(\Rightarrow x=\frac{1}{2}\) hoặc $x=\frac{5}{4}$ (đều thỏa mãn)
Vậy........
1) ĐK: \(x\ge\frac{3}{2}\)
pt \(\Leftrightarrow\frac{2x-2-\left(6x-9\right)}{\sqrt{2x-2}+\sqrt{6x-9}}=16x^2-28x-20x+35\)
\(\Leftrightarrow\frac{-4x+7}{\sqrt{2x-2}+\sqrt{6x-9}}=4x\left(4x-7\right)-5\left(4x-7\right)\)
\(\Leftrightarrow-\frac{4x-7}{\sqrt{2x-2}+\sqrt{6x-9}}=\left(4x-7\right)\left(4x-5\right)\)
\(\Leftrightarrow\left(4x-7\right)\left(\frac{1}{\sqrt{2x-2}+\sqrt{6x-9}}+4x-5\right)=0\)
\(\Leftrightarrow4x-7=0\Leftrightarrow x=\frac{7}{4}\) (nhận)
2) ĐK: \(2\le x\le4\)
pt \(\Leftrightarrow\sqrt{x-2}+\sqrt{a-x}=2\left(x^2-6x+9\right)+7x-19\)
\(\Leftrightarrow\sqrt{x-2}-\left(7x-20\right)+\sqrt{4-x}-1=2\left(x-3\right)^2\)
\(\Leftrightarrow\frac{x-2-\left(7x-20\right)^2}{\sqrt{x-2}+7x-20}+\frac{4-x-1}{\sqrt{4-x}+1}=2\left(x-3\right)^2\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(134-49x\right)}{\sqrt{x-2}+\left(7x-20\right)}+\frac{3-x}{\sqrt{4-x}+1}=2\left(x-3\right)^2\)
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\) (nhận)
2454 . 5424 .210 = (33 . 2 )24 . (23 . 3 )54 .210
= 672 . 6162 . 210
=72116640 \(⋮\) 72