Tìm \(x\) biết :
a) \(\sqrt{x-5}=3\)
b) \(\sqrt{x-10}=-2\)
c) \(\sqrt{2x-1}=\sqrt{5}\)
d) \(\sqrt{4-5x}=12\)
Tìm điều kiện có nghĩa:
1) \(\sqrt{2x^2}\)
2) \(\sqrt{-x}\)
3) \(\sqrt{-x^2-3}\)
4) \(\sqrt{x^2+2x+3}\)
5) \(\sqrt{-a^2+8a-16}\)
6) \(\sqrt[]{16x^2-25}\)
7) \(\sqrt{4x^2-49}\)
8) \(\sqrt{8-x^2}\)
9) \(\sqrt{x^2-12}\)
10) \(\sqrt{x^2+2x-3}\)
11) \(\sqrt{2x^2+5x+3}\)
12) \(\sqrt{\dfrac{4}{x-1}}\)
13) \(\sqrt{\dfrac{-1}{x-3}}\)
14) \(\sqrt{\dfrac{-3}{x+2}}\)
15) \(\sqrt{\dfrac{1}{2a-1}}\)
16) \(\sqrt{\dfrac{2}{3-2a}}\)
17) \(\sqrt{\dfrac{-1}{2a-5}}\)
18) \(\sqrt{\dfrac{-2}{3-5a}}\)
19) \(\sqrt{\dfrac{-a}{5}}\)
20) \(\dfrac{1}{\sqrt{-3a}}\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
Bài 1 GIẢI PHƯƠNG TRÌNH:
a) \(\sqrt{x-5}=\sqrt{3-x}\)
b) \(\sqrt{4-5x}=\sqrt{2-5x}\)
c) x2+4x+5=2\(\sqrt{2x+3}\)
d) \(\sqrt{x^2-2x+1}=\sqrt{4x^2-4x+1}\)
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
a) \(\sqrt{x-5}=\sqrt{3-x}\)
⇔\(\left(\sqrt{x-5}\right)^2=\left(\sqrt{3-x}\right)^2\)
⇔\(x-5=3-x\)
⇔\(x=4\)
b) \(\sqrt{4-5x}=\sqrt{2-5x}\)
⇔\(\left(\sqrt{4-5x}\right)^2=\left(\sqrt{2-5x}\right)^2\)
⇔\(4-5x=2-5x\)
⇔\(2=0\) (Vô lí)
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
a, \(\sqrt{x-5}=3\)
b, \(\sqrt{x-10}=-2\)
c, \(\sqrt{2x-1}=\sqrt{5}\)
d, \(\sqrt{4-5x}=12\)
\(a,\)
\(\sqrt{x-5}=3\)
\(\Leftrightarrow\)\(x-5=3^2\)
\(\Leftrightarrow\)\(x=14\)
\(b,\)
\(\sqrt{x-10}=-2\)
\(x\)không có giá trị ( vì \(\sqrt{x-10}\ge0\forall\))
\(c,\)
\(\sqrt{2x-1}=\sqrt{5}\)
\(\Leftrightarrow2x-1=\sqrt{5^2}\)
\(\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
\(d,\)
\(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=12^2\)
\(\Leftrightarrow5x=4-144\)
\(\Leftrightarrow5x=-140\)
\(\Leftrightarrow x=-\frac{140}{5}=-28\)
a, \(\sqrt{x-5}=3;ĐK:x-5\ge0\Leftrightarrow x\ge5\)
Ta có: \(\sqrt{x-5}=3\Leftrightarrow x-5=9\Leftrightarrow x=14\)
b, \(\sqrt{x-10}=-2;ĐK:x-10\ge0\Leftrightarrow x\ge10\)
Vì: \(\sqrt{x-10}\ge0\) nên không có giá trị nào của x để \(\sqrt{x-10}=-2\)
c, \(\sqrt{2x-1}=\sqrt{5};ĐK:2x-1\ge0\Leftrightarrow x\ge0,5\)
Ta có: \(\sqrt{2x-1}=\sqrt{5}\Leftrightarrow2x-1=5\)
\(\Leftrightarrow2x=6\Leftrightarrow x=3\)
d, \(\sqrt{4-5x}=12;ĐK:4-5x\ge0\Leftrightarrow x\le\frac{4}{5}\)
Ta có: \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\)
\(\Leftrightarrow-5x=140\Leftrightarrow x=-28\)
Giải pt:
a)\(\sqrt{\left(4-x\right).\left(6+x\right)}=x^2-2x-12\)
b)(x+1).(x+4)=5.\(\sqrt{x^2+5x+28}\)
c)x(x+5)=2.\(\sqrt[3]{x^2+5x-2}-2\)
d)3\(\sqrt{x}+\dfrac{3}{2\sqrt{3}}=2x+\dfrac{1}{2x}-7\)
b: \(\Leftrightarrow\left(x^2+5x+4\right)=5\sqrt{x^2+5x+28}\)
Đặt \(x^2+5x+4=a\)
Theo đề, ta có \(5\sqrt{a+24}=a\)
=>25a+600=a2
=>a=40 hoặc a=-15
=>x2+5x-36=0
=>(x+9)(x-4)=0
=>x=4 hoặc x=-9
c: \(\Leftrightarrow x^2+5x=2\sqrt[3]{x^2+5x-2}-2\)
Đặt \(x^2+5x=a\)
Theo đề, ta có: \(a=2\sqrt[3]{a}-2\)
\(\Leftrightarrow\sqrt[3]{8a}=a+2\)
=>(a+2)3=8a
=>\(a^3+6a^2+12a+8-8a=0\)
\(\Leftrightarrow a^3+6a^2+4a+8=0\)
Đến đây thì bạn chỉ cần bấm máy là xong
a)\(\sqrt{1-x}\left(x-3x^2\right)=x^3-3x^2+2x+6\)
b)\(x^2+x+12\sqrt{x+1}=36\)
c)\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
d)\(\sqrt{x^2+12}-3x=\sqrt{x^2+5}-5\)
e)\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)
f)\(4x^3-25x^2+43x+x\sqrt{3x-2}=22+\sqrt{3x-2}\)
g)\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)
h)\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)
i)\(\sqrt{1-3x}-\sqrt[3]{3x-1}=\left|6x-2\right|\)
k)\(\sqrt{2x^3+3x^2-1}=2x^2+2x-x^3-1\)
l)\(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)
\(\Rightarrow x=1\)
\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)
bạn làm nốt pần này nhá
Giải phương trình:
a. \(\sqrt{4-5x}=12\)
b. \(10-2\sqrt{2x+1}=4\)
c. \(5-\sqrt{x-1}=7\)
d. \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
e. \(\sqrt{x+1}+10=2\sqrt{x+1}-2\)
f. \(\sqrt{16x+32}-5\sqrt{x+2}=-2\)
a. ĐKXĐ: \(4-5x\ge0\) \(\Leftrightarrow-5x\ge-4\Leftrightarrow5x\le4\Leftrightarrow x\le\dfrac{4}{5}\)
\(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=2\sqrt{3}\)
\(\Leftrightarrow-5x=-4-2\sqrt{3}\)
\(\Leftrightarrow x=\dfrac{-4-2\sqrt{3}}{-5}\)
\(\Leftrightarrow x=\dfrac{4+2\sqrt{3}}{5}\left(KTMĐKXĐ\right)\)
Vậy x không tồn tại
b. \(10-2\sqrt{2x+1}=4\) (1)
\(ĐKXĐ:2x+1\ge0\Leftrightarrow2x\ge-1\Leftrightarrow x\ge-\dfrac{1}{2}\)
(1) => \(-2\sqrt{2x+1}=-6\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=\sqrt{3}\)
\(\Leftrightarrow2x=\sqrt{3}-1\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}-1}{2}\left(TMĐKXĐ\right)\)
c. \(5-\sqrt{x-1}=7\) (1)
ĐKXĐ: \(x-1\ge0\Leftrightarrow x\ge1\)
(1) <=> \(-\sqrt{x-1}=2\) (vô lí)
Vậy không tồn tại x
bài kia làm sai rùi:
a. \(\sqrt{4-5x}=12\) (1)
ĐKXĐ: \(4-5x\ge0\Leftrightarrow x\le\dfrac{4}{5}\)
\(\Leftrightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
\(\Leftrightarrow x=-28\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là \(S=\left\{-28\right\}\)
b. \(10-2\sqrt{2x+1}=4\) (1)
ĐKXĐ: \(2x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2\sqrt{2x+1}=6\)
\(\Leftrightarrow\sqrt{2x+1}=3\)
\(\Leftrightarrow2x+1=9\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là: \(S=\left\{4\right\}\)
c. Ở dưới làm đúng rồi
d. \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\) (1)
ĐKXĐ: \(3x\ge0\Leftrightarrow x\ge0\)
(1) \(\Leftrightarrow10+\sqrt{3x}=\left(2+\sqrt{6}\right)^2\)
\(\Leftrightarrow10+\sqrt{3x}=10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=-10+10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
\(\Leftrightarrow x=32\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là: \(S=\left\{32\right\}\)
e. \(\sqrt{x+1}+10=2\sqrt{x+1}-2\) (1)
ĐKXĐ: \(x+1\ge0\Leftrightarrow x\ge-1\)
\(\left(1\right)\Leftrightarrow\sqrt{x+1}-2\sqrt{x+1}=-10-2\)
\(\Leftrightarrow-\sqrt{x+1}=-12\)
\(\Leftrightarrow\sqrt{x+1}=12\)
\(\Leftrightarrow x+1=144\)
\(\Leftrightarrow x=143\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là \(S=\left\{143\right\}\)
f. \(\sqrt{16x+32}-5\sqrt{x+2}=-2\) (1)
ĐKXĐ: \(\left[{}\begin{matrix}\sqrt{16x+32\ge0}\\\sqrt{x+2\ge0}\end{matrix}\right.\left[{}\begin{matrix}x\ge-2\\x\ge-2\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{16\left(x+2\right)}-5\sqrt{x+2}=-2\)
\(\Leftrightarrow4\sqrt{x+2}-5\sqrt{x+2}=-2\)
\(\Leftrightarrow-\sqrt{x+2}=-2\)
\(\Leftrightarrow\sqrt{x+2}=2\)
\(\Leftrightarrow x+2=4\)
\(\Leftrightarrow x=2\left(TMĐKXĐ\right)\)
Vậy phương trình có nghiệm là \(S=\left\{2\right\}\)
a)\(\sqrt{4-5x}=12\) tìm x
b)\(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
c)\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
a) Ta có: \(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
hay x=-28
b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
hay x=32
c) Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)