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Thiên Thư mk cx hk lp 7 nek
a\ \(\sqrt{x^2-4x+4}=6\)
\(x^2-4x+4=6^2=36\)
\(x\left(x-4\right)=32\)
ta có \(32=8.4=\left(-8\right)\left(-4\right)\)
\(\Rightarrow x\in\left\{8;-4\right\}\)
b)\(\sqrt{2x+5}=2x-1\)
\(2x+4=4x^2-4x\)
\(2\left(x+2\right)=4x\left(4x-1\right)\)
\(........................\)
e bí mất r a ạ
ĐKXĐ:....
\(\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}\)
\(\Rightarrow4-\sqrt{1-x}=2-x\)
\(\Rightarrow\sqrt{1-x}=2+x\)
\(\Rightarrow1-x=4+4x+x^2\)
\(\Rightarrow1-x-4-4-x^2=0\)
\(\Rightarrow x^2+x+7=0\)
Đến đây dễ rồi làm nốt nha bạn !
ĐKXĐ:....
\sqrt{4-\sqrt{1-x}}=\sqrt{2-x}4−1−x=2−x
\Rightarrow4-\sqrt{1-x}=2-x⇒4−1−x=2−x
\Rightarrow\sqrt{1-x}=2+x⇒1−x=2+x
\Rightarrow1-x=4+4x+x^2⇒1−x=4+4x+x2
\Rightarrow1-x-4-4-x^2=0⇒1−x−4−4−x2=0
\Rightarrow x^2+x+7=0⇒x2+x+7=0
Đến đây dễ rồi làm nốt nha bạn !
\(a,\sqrt{x-1}+\sqrt{9-x}=4\)
\(ĐKXĐ:1\le x\le9\)
\(\sqrt{x-1}=4-\sqrt{9-x}\)
\(x-1=16-8\sqrt{9-x}+9-x\)
\(26-8\sqrt{9-x}-2x=0\)
\(13-4\sqrt{9-x}-x=0\)
\(9-x-4\sqrt{9-x}+4=0\)
\(\left(\sqrt{9-x}-2\right)^2=0\)
\(\sqrt{9-x}=2\)
\(9-x=4\)
\(x=5\left(TM\right)\)
\(\sqrt{2x-1}+\sqrt{x+4}=6\)
\(ĐKXĐ:x\ge\frac{1}{2}\)
\(x+4=36-12\sqrt{2x-1}+2x-1\)
\(x+4=35-12\sqrt{2x-1}+2x\)
\(31-12\sqrt{2x-1}+x=0\)
\(\left(31+x\right)^2=\left(12\sqrt{2x-1}\right)^2\)
\(961+62x+x^2=144\left(2x-1\right)\)
\(961+62x+x^2=288x-144\)
\(x^2-226x+1105=0\)
\(\sqrt{\Delta}=216\)
\(x_1=\frac{226+216}{2}=221\left(TM\right)\)
\(x_2=\frac{226-216}{2}=5\left(TM\right)\)
................................................. tui ko bít
a) ĐK: \(x\ge\frac{1}{2}\).
\(\sqrt{2x-1}+\sqrt{x+4}=6\)
\(\Leftrightarrow\sqrt{2x-1}-3+\sqrt{x+4}-3=0\)
\(\Leftrightarrow\frac{2x-1-9}{\sqrt{2x-1}+3}+\frac{x+4-9}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}\right)=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\).
b) ĐK: \(x\ge\frac{1}{2}\).
\(\sqrt{x+3}-\sqrt{2x-1}=1\)
\(\Leftrightarrow\sqrt{x+3}-2+1-\sqrt{2x-1}=0\)
\(\Leftrightarrow\frac{x+3-4}{\sqrt{x+3}+2}+\frac{1-\left(2x-1\right)}{1+\sqrt{2x-1}}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{\sqrt{x+3}+2}-\frac{2}{1+\sqrt{2x-1}}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\\frac{1}{\sqrt{x+3}+2}=\frac{2}{1+\sqrt{2x-1}}\left(1\right)\end{cases}}\)
\(\left(1\right)\Leftrightarrow2\sqrt{x+3}+4=1+\sqrt{2x-1}\)
Có \(4>1,2\sqrt{x+3}=\sqrt{4x+12}>\sqrt{2x-1}\)
do đó phương trình \(\left(1\right)\)vô nghiệm.
a) ĐK : x >= 1/2
\(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x+4}-3\right)=0\)
\(\Leftrightarrow\frac{2x-1-9}{\sqrt{2x-1}+3}+\frac{x+4-9}{\sqrt{x+4}+3}=0\)
\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}\right)=0\)(1)
Dễ thấy với x >= 1/2 thì \(\frac{2}{\sqrt{2x-1}+3}+\frac{1}{\sqrt{x+4}+3}>0\)
nên (1) <=> x - 5 = 0 <=> x = 5 (tm)
Vậy phương trình có nghiệm x = 5