Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\cdot\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)

\(< =>2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)

\(< =>2\cdot3\sqrt{\left(x-3\right)}-\frac{1}{5}.5\sqrt{\left(x-3\right)}-\frac{1}{7}.7\sqrt{\left(x-3\right)}=20\) \(đk:x\ge0\)

\(< =>6\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}-\sqrt{\left(x-3\right)}=20\)

\(< =>\sqrt{\left(x-3\right)}\left(6-1-1\right)=20\)

\(< =>4\sqrt{\left(x-3\right)}=20\)

\(< =>\sqrt{\left(x-3\right)}=5\)

\(< =>x-3=25\)

\(< =>x=28\left(tm\right)\)

đề =

 \(2\sqrt{9\left(x-3\right)}-\frac{1}{5}\sqrt{25\left(x-3\right)}-\frac{1}{7}\sqrt{49\left(x-3\right)}=20\)

=>\(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

=>\(4\sqrt{x-3}=20\)

=>\(\sqrt{x-3}=5\)

=>\(x-3=25\)

=>\(x=28\)

\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{3;-1\right\}\)

\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)

Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)

\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)

Vậy \(S=\left\{28\right\}\)

1. ĐKXĐ: \(x\ge3\)

\(2\sqrt{9x-27}-\frac{1}{5}\sqrt{25x-75}-\frac{1}{7}\sqrt{49x-147}=20\)

⇔ \(6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

⇔ \(4\sqrt{x-3}=20\)

⇔ \(\sqrt{x-3}=5\)

⇔ \(x-3=25\)

⇔ \(x=28\left(TMĐKXĐ\right)\)

Vậy....

2. ĐKXĐ: \(x\ge0\)

\(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-7=\frac{1}{2}\sqrt{5x}\)

⇔ \(\frac{3}{2}\sqrt{5x}+\sqrt{5x}-\frac{1}{2}\sqrt{5x}=7\)

⇔ \(2\sqrt{5x}=7\)

⇔ \(\sqrt{5x}=\frac{7}{2}\)

⇔ \(5x=\frac{49}{4}\)

⇔ \(x=\frac{49}{20}\left(TMĐKXĐ\right)\)

Vậy...

a) Ta có: \(2\sqrt{9x-27}-\dfrac{1}{5}\sqrt{25x-75}-\dfrac{1}{7}\sqrt{49x-147}=20\)

\(\Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\)

\(\Leftrightarrow4\sqrt{x-3}=20\)

\(\Leftrightarrow x-3=25\)

hay x=28

b) Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow x+2=9\)

hay x=7

1. Ta có: \(25x^2+...+49y^2=\left(5x\right)^2+...+\left(7y\right)^2\)

Áp dụng HĐT thứ 1: \(\left(5x+7y\right)^2=\left(5x\right)^2+2.5x.7y+\left(7y\right)^2\)

Vậy biểu thức cần điền vào chỗ trống là 70xy.

2. Ta có: \(36x^2-12xy+...=\left(6x\right)^2-2.6x.2y+...\)

Áp dụng HĐT thứ 2: \(\left(6x-2y\right)^2=\left(6x\right)^2-2.6x.2y+\left(2y\right)^2\)

Vậy biểu thức cần điền vào chỗ trống là 4y².

3. Ta có: \(...+30xy+25y^2=...+2.6x.5y+\left(5y\right)^2\)

Áp dụng HĐT thứ 1: \(\left(6x+5y\right)^2=\left(6x\right)^2+2.6x.5y+\left(5y\right)^2\)

Vậy biểu thức cần điền vào chỗ trống là 36x².

4. \(8x^3+1=\left(2x\right)^3+1^3\)

\(=\left(2x+1\right)\left[\left(2x\right)^2-2x.1+1^2\right]\)

\(=\left(2x+1\right)\left(4x^2-2x+1\right)\)

5. \(27x^3-8y^3=\left(3x\right)^3-\left(2y\right)^3\)

\(=\left(3x-2y\right)\left[\left(3x\right)^2+3x.2y+\left(2y\right)^2\right]\)

\(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)

a) 18*76+24*18-150=

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